Question 1

Draw a quadrilateral in the Cartesian plane, whose vertices are (-4, 5), (0, 7), (5, -5) and (-4, -2). Also, find its area.

Accepted Answer

121/2 square unit.. Plot the points, then split the quadrilateral into two triangles. Mark A(-4,5), B(0,7), C(5,-5), D(-4,-2) and join them in that order. A rough sketch on graph paper helps here: A and D both lie on the vertical line x=-4, so AD is the left-hand side, B is the top vertex and C the bottom-right one, and ABCD closes up without crossing itself. Join the diagonal AC. It cuts the quadrilateral into ABC and ACD, so ar (ABCD)= ar ( ABC)+ ar ( ACD). For a triangle with vertices (x 1,y 1),(x 2,y 2),(x 3,y 3), Area =1/2 x 1(y 2-y 3)+x 2(y 3-y 1)+x 3(y 1-y 2). For ABC with A(-4,5),B(0,7),C(5,-5): ar ( ABC)=1/2 -4(7+5)+0(-5-5)+5(5-7) =1/2 -48+0-10 =58/2=29. For ACD with A(-4,5),C(5,-5),D(-4,-2): ar ( ACD)=1/2 -4(-5+2)+5(-2-5)+(-4)(5+5) =1/2 12-35-40 =63/2=31.5. Adding, 29+31.5=60.5. Area of the quadrilateral =121/2=60.5 square units.

Question 2

The base of an equilateral triangle with side 2 a lies along the y -axis such that the mid-point of the base is at the origin. Find vertices of the triangle.

Accepted Answer

(0, a),(0,-a) and (-√ 3 a, 0) or (0, a),(0,-a), and (√ 3 a, 0). Use the symmetry of the base about the origin. The base lies along the y -axis with its mid-point at the origin, and its length is 2a. So the two base vertices are at distance a on either side of the origin along the y -axis: B(0,a) and C(0,-a). In an equilateral triangle the third vertex A lies on the perpendicular bisector of the base. The perpendicular bisector of BC is the x -axis, so A=(h,0) for some h. Since AB=2a, by the distance formula √ (h-0)^2+(0-a)^2 =2a ⇒ h^2+a^2=4a^2 ⇒ h^2=3a^2 ⇒ h=±√ 3 a. (Equivalently, the altitude of an equilateral triangle of side 2a is (√3)/2(2a)=√3 a.) Either sign gives a valid triangle — one to the right of the y -axis, one to the left. The vertices are (0,a), (0,-a) and (√ 3 a, 0); or (0,a), (0,-a) and (-√ 3 a, 0).

Question 3

Find the distance between P (x 1, y 1) and Q (x 2, y 2) when: (i) PQ is parallel to the y -axis, (ii) PQ is parallel to the x -axis.

Accepted Answer

(i) y 2-y 1, (ii) x 2-x 1. Distance formula, with one coordinate common. In general PQ=√ (x 2-x 1)^2+(y 2-y 1)^2. (i) PQ parallel to the y -axis. A line parallel to the y -axis is vertical, so both points have the same abscissa: x 1=x 2. Then PQ=√ 0+(y 2-y 1)^2 = y 2-y 1. (ii) PQ parallel to the x -axis. A line parallel to the x -axis is horizontal, so both points have the same ordinate: y 1=y 2. Then PQ=√ (x 2-x 1)^2+0 = x 2-x 1. The modulus is needed because a distance is never negative. (i) PQ= y 2-y 1 (ii) PQ= x 2-x 1.

Question 4

Find a point on the x -axis, which is equidistant from the points (7,6) and (3,4).

Accepted Answer

(15/2, 0). Distance formula, with the unknown point written as (x,0). Every point of the x -axis has ordinate 0, so let the required point be P(x,0). Let A(7,6) and B(3,4). "Equidistant" means PA=PB, and since both are non-negative we may square: PA^2=PB^2 (x-7)^2+(0-6)^2=(x-3)^2+(0-4)^2 x^2-14x+49+36=x^2-6x+9+16 -14x+85=-6x+25 -8x=-60 ⇒ x=15/2. Check: with x=15/2, PA^2=(15/2-7)^2+36= 14+36=145/4 and PB^2=(15/2-3)^2+16=81/4+16=145/4. Equal, as required. The point is (15/2, 0).

Question 5

Find the slope of a line, which passes through the origin, and the mid-point of the line segment joining the points P (0,-4) and B (8,0).

Accepted Answer

Mid-point formula, then the slope formula. The mid-point of the segment joining (0,-4) and (8,0) is ((0+8)/2, (-4+0)/2)=(4,-2). The required line passes through the origin (0,0) and this mid-point (4,-2). For two points (x 1,y 1) and (x 2,y 2) with x 1≠ x 2, m=y 2-y 1/x 2-x 1=-2-0/4-0=-1/2. Slope =-1/2.

Question 6

Without using the Pythagoras theorem, show that the points (4, 4), (3,5) and (-1, -1) are the vertices of a right angled triangle.

Accepted Answer

Slope test for perpendicularity (no distances used). Let A(4,4), B(3,5), C(-1,-1). Two lines with slopes m 1,m 2 are perpendicular exactly when m 1m 2=-1, so it is enough to find the three slopes. m AB=5-4/3-4=1/-1=-1, m AC=-1-4/-1-4=-5/-5=1, m BC=-1-5/-1-3=-6/-4=3/2. Now m AB× m AC=(-1)(1)=-1, so AB⊥ AC, i.e. the angle at A(4,4) is a right angle. Also m AB≠ m AC≠ m BC, so the three points are not collinear and do form a triangle. The points are the vertices of a right-angled triangle, right-angled at (4,4).

Question 7

Find the slope of the line, which makes an angle of 30^° with the positive direction of y -axis measured anticlockwise.

Accepted Answer

-√ 3. Convert the given angle into the inclination with the positive x -axis. The slope of a line is m= tan θ, where θ is the inclination measured from the positive x -axis, anticlockwise. The positive y -axis is already at 90^° from the positive x -axis. Turning a further 30^° anticlockwise gives θ=90^°+30^°=120^°. Hence m= tan 120^°= tan (180^°-60^°)=- tan 60^°=-√ 3. Slope =-√ 3.

Question 8

Without using distance formula, show that points (-2, -1), (4, 0), (3, 3) and (-3, 2) are the vertices of a parallelogram.

Accepted Answer

Slope test: a quadrilateral with both pairs of opposite sides parallel is a parallelogram. Let A(-2,-1), B(4,0), C(3,3), D(-3,2), taken in that order. Two lines are parallel exactly when their slopes are equal, so compute the slopes of the four sides. m AB=(0-(-1))/(4-(-2))=1/6, m DC=3-2/(3-(-3))=1/6, m BC=3-0/3-4=3/-1=-3, m AD=(2-(-1))/(-3-(-2))=3/-1=-3. So m AB=m DC⇒ AB∥ DC, and m BC=m AD⇒ BC∥ AD. Since m AB≠ m BC, adjacent sides are not along the same line, so ABCD is a genuine quadrilateral. Both pairs of opposite sides are parallel. Hence ABCD is a parallelogram.

Question 9

Find the angle between the x -axis and the line joining the points (3,-1) and (4,-2).

Accepted Answer

135^°. Slope of the line = tan of its inclination with the x -axis. Slope of the line joining (3,-1) and (4,-2): m=(-2-(-1))/4-3=-1/1=-1. If θ is the angle the line makes with the positive direction of the x -axis, then tan θ=m, so tan θ=-1. Taking θ in [0^°,180^°), the inclination of a line, θ=180^°-45^°=135^°. (The acute angle the line makes with the x -axis is therefore 45^°.) The angle is 135^°.

Question 10

The slope of a line is double of the slope of another line. If tangent of the angle between them is 1/3, find the slopes of the lines.

Accepted Answer

1 and 2, or 1/2 and 1, or - 1 and -2, or - 1/2 and - 1. Angle-between-two-lines formula. Let the slopes be m and 2m. If θ is the angle between two lines of slopes m 1,m 2, then tan θ= m 2-m 1/(1+m 1m 2). Here tan θ= 2m-m/(1+m· 2m) = m/(1+2m^2) =1/3. So m/(1+2m^2)=± 13. Case +: 3m=1+2m^2⇒ 2m^2-3m+1=0⇒(2m-1)(m-1)=0⇒ m= 12 or m=1. Case -: -3m=1+2m^2⇒ 2m^2+3m+1=0⇒(2m+1)(m+1)=0⇒ m=- 12 or m=-1. Each value of m gives the pair (m,2m); substituting back, every one of them gives m/(1+2m^2) = 13. The slopes are 1 and 2; or 12 and 1; or -1 and -2; or - 12 and -1.

Question 11

A line passes through (x 1, y 1) and (h, k). If slope of the line is m, show that k-y 1=m(h-x 1).

Accepted Answer

Slope formula for a line through two known points. If a line passes through (x 1,y 1) and (x 2,y 2) with x 1≠ x 2, its slope is m=y 2-y 1/x 2-x 1. Here the two points are (x 1,y 1) and (h,k), so m=k-y 1/h-x 1. Multiplying both sides by h-x 1 (which is non-zero, since a slope m exists only when the line is not vertical): k-y 1=m(h-x 1). Hence k-y 1=m(h-x 1), as required.

Question 12

Write the equations for the x -and y -axes.

Accepted Answer

y=0 and x=0. Read off the defining property of each axis. The x -axis. Every point on the x -axis has ordinate 0, and no other point does. So a point (x,y) lies on the x -axis if and only if y=0. (Consistently: the x -axis passes through (0,0) with slope 0, so y-0=0(x-0), i.e. y=0.) The y -axis. Every point on the y -axis has abscissa 0, and no other point does. So a point (x,y) lies on the y -axis if and only if x=0. (This line is vertical, so it has no slope and cannot be written in slope form.) Equation of the x -axis: y=0. Equation of the y -axis: x=0.

Question 13

Passing through the point (-4,3) with slope 1/2.

Accepted Answer

x-2 y+10=0. Point-slope form. A line of slope m through (x 0,y 0) has equation y-y 0=m(x-x 0). Here (x 0,y 0)=(-4,3) and m= 12: y-3=1/2(x-(-4))=1/2(x+4). Multiply through by 2: 2y-6=x+4 ⇒ x-2y+10=0. Check: at (-4,3), -4-6+10=0. ✓ x-2y+10=0.

Question 14

Passing through (0,0) with slope m.

Accepted Answer

y=m x. Point-slope form with the point at the origin. A line of slope m through (x 0,y 0) has equation y-y 0=m(x-x 0). Here (x 0,y 0)=(0,0), so y-0=m(x-0). y=mx.

Question 15

Passing through (2,2 √ 3 ) and inclined with the x -axis at an angle of 75°.

Accepted Answer

(√ 3 +1) x-(√ 3 -1) y=4(√ 3 -1). Point-slope form, with the slope obtained from the inclination. The slope is m= tan 75^°. Write 75^°=45^°+30^° and use tan (A+B)=( tan A+ tan B)/(1- tan A tan B): tan 75^°=( tan 45^°+ tan 30^°)/(1- tan 45^° tan 30^°)=(1+1/(√3))/(1-1/(√3))=(√3+1)/(√3-1)=((√3+1)^2)/((√3-1)(√3+1))=(4+2√3)/2=2+√3. Now apply y-y 0=m(x-x 0) with (x 0,y 0)=(2, 2√3): y-2√3=(2+√3)(x-2) y-2√3=(2+√3)x-4-2√3 y=(2+√3)x-4. (2+√ 3 )x-y=4.

Question 16

Intersecting the x -axis at a distance of 3 units to the left of origin with slope -2.

Accepted Answer

2 x+y+6=0. Point-slope form, after naming the point of intersection with the x -axis. "Intersecting the x -axis at a distance of 3 units to the left of the origin" means the line passes through (-3,0). The slope is m=-2. Using y-y 0=m(x-x 0): y-0=-2(x-(-3))=-2(x+3) y=-2x-6 2x+y+6=0. Check: at (-3,0), -6+0+6=0. ✓ 2x+y+6=0.

Question 17

Intersecting the y -axis at a distance of 2 units above the origin and making an angle of 30° with positive direction of the x -axis.

Accepted Answer

x-√ 3 y+2 √ 3 =0. Slope-intercept form. A line of slope m with y -intercept c has equation y=mx+c. The line cuts the y -axis 2 units above the origin, so c=2. It makes 30^° with the positive x -axis, so m= tan 30^°=1/(√3). Hence y=1/(√3)x+2. Multiplying by √3: √3 y=x+2√3. x-√ 3 y+2√ 3 =0.

Question 18

Passing through the points (-1, 1) and (2, -4).

Accepted Answer

5 x+3 y+2=0. Two-point form. The line through (x 1,y 1) and (x 2,y 2) is y-y 1=y 2-y 1/x 2-x 1(x-x 1). With (x 1,y 1)=(-1,1) and (x 2,y 2)=(2,-4), the slope is m=-4-1/(2-(-1))=-5/3. So y-1=-5/3(x+1) 3y-3=-5x-5 5x+3y+2=0. Check: at (-1,1), -5+3+2=0; at (2,-4), 10-12+2=0. ✓ 5x+3y+2=0.

Question 19

The vertices of Δ PQR are P (2, 1), Q (-2, 3) and R (4, 5). Find equation of the median through the vertex R.

Accepted Answer

3 x-4 y+8=0. Median through a vertex. A median joins a vertex to the mid-point of the opposite side, so the median through R joins R to the mid-point of PQ. Mid-point of PQ, with P (2, 1) and Q (-2, 3): M =((2+(-2))/2, (1+3)/2)=(0, 2). Slope of the line through M (0, 2) and R (4, 5): m=5-2/4-0=3/4. Using the point-slope form y-y 1=m(x-x 1) with the point M (0, 2): y-2=3/4(x-0) ⇒ 4y-8=3x. Both P and Q are checked against the result only through M; substituting R (4, 5) gives 3(4)-4(5)+8=0, so R lies on it as required. The median through R is 3x-4y+8=0.

Question 20

Find the equation of the line passing through (-3,5) and perpendicular to the line through the points (2, 5) and (-3, 6).

Accepted Answer

5 x-y+20=0. Point-slope form with the perpendicularity condition. Two lines are perpendicular when the product of their slopes is -1. Slope of the line through (2, 5) and (-3, 6): m 1=6-5/-3-2=1/-5=-1/5. The required line is perpendicular to it, so its slope is m=-1/m 1=5. It passes through (-3,5), so by the point-slope form: y-5=5 (x-(-3) )=5(x+3) ⇒ y-5=5x+15. The required line is 5x-y+20=0.

Question 21

A line perpendicular to the line segment joining the points (1,0) and (2,3) divides it in the ratio 1: n. Find the equation of the line.

Accepted Answer

(1+n) x+3(1+n) y=n+11. Section formula, then point-slope form. Let the line meet the segment joining A (1, 0) and B (2, 3) at the point C that divides AB in the ratio 1:n. By the section formula, the point dividing the join of (x 1,y 1) and (x 2,y 2) internally in the ratio m 1:m 2 is ((m 1x 2+m 2x 1)/(m 1+m 2),(m 1y 2+m 2y 1)/(m 1+m 2)). With m 1:m 2=1:n: C =((1· 2+n· 1)/(1+n), (1· 3+n· 0)/(1+n))=((n+2)/(n+1), 3/(n+1)). Slope of AB is 3-0/2-1=3, so the slope of the perpendicular line is -1/3. Point-slope form through C: y-3/(n+1)=-1/3(x-(n+2)/(n+1)). Multiplying through by 3(n+1): 3(n+1)y-9=-(n+1)x+(n+2), (1+n)x+3(1+n)y=n+11. The required line is (1+n)x+3(1+n)y=n+11.

Question 22

Find the equation of a line that cuts off equal intercepts on the coordinate axes and passes through the point (2, 3).

Accepted Answer

x+y=5. Intercept form. A line with intercepts a on the x -axis and b on the y -axis (both non-zero) is x/a+y/b=1. The intercepts are equal, so b=a and the line is x/a+y/a=1 ⇒ x+y=a. It passes through (2, 3), so a=2+3=5. Check: 2+3=5. ✓ The required line is x+y=5.

Question 23

Find equation of the line passing through the point (2,2) and cutting off intercepts on the axes whose sum is 9.

Accepted Answer

x+2 y-6=0,2 x+y-6=0. Intercept form with a constraint. Let the intercepts be a and b, so the line is x/a+y/b=1. Given a+b=9. Since the line passes through (2, 2), 2/a+2/b=1 ⇒ 2(a+b)=ab ⇒ ab=2(9)=18. So a and b are the roots of t^2-(a+b)t+ab=0 ⇒ t^2-9t+18=0 ⇒ (t-3)(t-6)=0, giving t=3 or t=6. Hence (a,b)=(3,6) or (6,3). For (a,b)=(3,6): x/3+y/6=1⇒ 2x+y=6. For (a,b)=(6,3): x/6+y/3=1⇒ x+2y=6. Both pass through (2, 2): 2(2)+2=6 and 2+2(2)=6. ✓ The lines are 2x+y-6=0 and x+2y-6=0.

Question 24

Find equation of the line through the point (0,2) making an angle (2 π)/3 with the positive x -axis. Also, find the equation of line parallel to it and crossing the y -axis at a distance of 2 units below the origin.

Accepted Answer

√ 3 x+y-2=0 and √ 3 x+y+2=0. Slope-intercept form. The slope of a line making an angle θ with the positive x -axis is m= tan θ. Here θ=(2π)/3=120^°, so m= tan 120^°=- tan 60^°=-√ 3. The line passes through (0, 2), which is its y -intercept, so by y=mx+c: y=-√ 3 x+2 ⇒ √ 3 x+y-2=0. A parallel line has the same slope -√ 3. Crossing the y -axis 2 units below the origin means it passes through (0,-2), so c=-2: y=-√ 3 x-2 ⇒ √ 3 x+y+2=0. The line is √ 3 x+y-2=0, and the parallel line is √ 3 x+y+2=0.

Question 25

The perpendicular from the origin to a line meets it at the point (-2, 9), find the equation of the line.

Accepted Answer

2 x-9 y+85=0. Perpendicular from the origin. Let the foot of the perpendicular be A (-2, 9). The segment OA is perpendicular to the required line, so their slopes multiply to -1. Slope of OA, with O (0, 0) and A (-2, 9): m OA =9-0/-2-0=-9/2. Hence the slope of the line is m=- 1 m OA =2/9. The line passes through A (-2, 9), so by the point-slope form: y-9=2/9(x+2) ⇒ 9y-81=2x+4. The required line is 2x-9y+85=0.

Question 26

The length L (in centimetre) of a copper rod is a linear function of its Celsius temperature C. In an experiment, if L =124.942 when C =20 and L =125.134 when C =110, express L in terms of C.

Accepted Answer

L =.192/90( C -20)+124.942. Two-point form for a linear relation. Since L is a linear function of C, the points ( C, L ) lie on a straight line through (20, 124.942) and (110, 125.134). Slope: m=125.134-124.942/110-20=0.192/90=4/1875≈ 0.002133. Two-point (point-slope) form using (20, 124.942): L -124.942=4/1875( C -20), L =124.942+4/1875 C -80/1875=4/1875 C +124.8993 3. Check: at C =110, 4/1875(110)+124.89933…=0.234 6 +124.89933…=125.134. ✓ L =4/1875 C +124.8993…, i.e. L =124.942+4/1875( C -20)≈ 0.002133 C +124.899.

Question 27

The owner of a milk store finds that, he can sell 980 litres of milk each week at Rs 14/litre and 1220 litres of milk each week at Rs 16/litre. Assuming a linear relationship between selling price and demand, how many litres could he sell weekly at Rs 17/litre?

Accepted Answer

1340 litres.. Two-point form of a linear model. Let x be the selling price in rupees per litre and y the weekly demand in litres. The data give the points (14, 980) and (16, 1220). Slope: m=1220-980/16-14=240/2=120. Point-slope form through (14, 980): y-980=120(x-14) ⇒ y=120x-1680+980=120x-700. Check: at x=16, y=120(16)-700=1920-700=1220. ✓ At x=17: y=120(17)-700=2040-700=1340. He could sell 1340 litres of milk weekly at Rs 17/litre.

Question 28

P (a, b) is the mid-point of a line segment between axes. Show that equation of the line is x/a+y/b=2.

Accepted Answer

Intercept form with the mid-point formula. Picture the line cutting the axes so that it forms a right triangle with them: it meets the x -axis at A (p,0) and the y -axis at B (0,q), and AB is the "segment between the axes". P (a,b) is the mid-point of AB, so by the mid-point formula ((p+0)/2, (0+q)/2)=(a,b) ⇒ p/2=a, q/2=b, p=2a, q=2b. The intercept form of the line with intercepts p and q is x/p+y/q=1, so x/2a+y/2b=1. Multiplying both sides by 2: x/a+y/b=2. Hence the equation of the line is x/a+y/b=2, as required.

Question 29

Point R (h, k) divides a line segment between the axes in the ratio 1: 2. Find equation of the line.

Accepted Answer

2 k x+h y=3 k h.. Intercept form with the section formula. Picture the line cutting the axes at A (p,0) on the x -axis and B (0,q) on the y -axis; AB is the "segment between the axes", and R lies on it nearer to A. R (h,k) divides AB in the ratio 1:2, so by the section formula (h,k)=((1· 0+2· p)/(1+2), (1· q+2· 0)/(1+2))=(2p/3, q/3). Therefore h=2p/3 ⇒ p=3h/2, k=q/3 ⇒ q=3k. Intercept form x/p+y/q=1 gives x/(3h/2)+y/3k=1 ⇒ 2x/3h+y/3k=1. Multiplying by 3: 2x/h+y/k=3. Check with h=3, k=2: then p= 92, q=6, and R =(2/3· 92, 63)=(3,2). ✓ The required line is 2x/h+y/k=3, that is, 2kx+hy=3hk.

Question 30

By using the concept of equation of a line, prove that the three points (3,0), (-2, -2) and (8,2) are collinear.

Accepted Answer

Equation of a line through two points. Three points are collinear if the third point satisfies the equation of the line through the first two. Line through A (3, 0) and B (-2, -2). Slope: m=-2-0/-2-3=-2/-5=2/5. Point-slope form through A (3, 0): y-0=2/5(x-3) ⇒ 5y=2x-6 ⇒ 2x-5y-6=0. Now substitute C (8, 2) into the left-hand side: 2(8)-5(2)-6=16-10-6=0. Since C satisfies the equation, it lies on the line AB. Hence the points (3, 0), (-2, -2) and (8, 2) are collinear.

Question 31

Reduce the following equations into slope - intercept form and find their slopes and the y - intercepts. (i) x+7 y=0, (ii) 6 x+3 y-5=0, (iii) y=0.

Accepted Answer

(i) y=-1/7 x+0,-1/7, 0; (ii) y=-2 x+5/3,-2, 5/3; (iii) y=0 x+0,0,0. Slope-intercept form. Write each equation as y=mx+c; then m is the slope and c is the y -intercept. (i) x+7y=0 7y=-x ⇒ y=-1/7x+0. Slope =-1/7, y -intercept =0. (ii) 6x+3y-5=0 3y=-6x+5 ⇒ y=-2x+5/3. Slope =-2, y -intercept =5/3. (iii) y=0 y=0· x+0. Slope =0, y -intercept =0. (This is the x -axis itself.) Slopes and y -intercepts: (i) -1/7 and 0; (ii) -2 and 5/3; (iii) 0 and 0.

Question 32

Reduce the following equations into intercept form and find their intercepts on the axes. (i) 3 x+2 y-12=0, (ii) 4 x-3 y=6, (iii) 3 y+2=0.

Accepted Answer

(i) x/4+y/6=1,4,6; (ii) x/(3/2)+y/-2=1, 3/2,-2; (iii) y=-2/3, intercept with y -axis =-2/3 and no intercept with x -axis.. Intercept form. Move the constant to the right and divide through so that the right-hand side becomes 1, giving x/a+y/b=1; then a is the x -intercept and b the y -intercept. (i) 3x+2y-12=0⇒ 3x+2y=12. Dividing by 12: 3x/12+2y/12=1 ⇒ x/4+y/6=1. x -intercept =4, y -intercept =6. (ii) 4x-3y=6. Dividing by 6: 4x/6-3y/6=1 ⇒ x/(3/2)+y/-2=1. x -intercept =3/2, y -intercept =-2. (iii) 3y+2=0⇒ 3y=-2. Dividing by -2: y/(-2/3)=1. y -intercept =-2/3. There is no x term, so the line is parallel to the x -axis and has no x -intercept. Intercepts: (i) 4 and 6; (ii) 3/2 and -2; (iii) no x -intercept and -2/3.

Question 33

Find the distance of the point (-1,1) from the line 12(x+6)=5(y-2).

Accepted Answer

5 units. Perpendicular distance formula. The distance of (x 1,y 1) from Ax+By+C=0 is d= Ax 1+By 1+C √ A^2+B^2. First put the line in general form: 12(x+6)=5(y-2) ⇒ 12x+72=5y-10 ⇒ 12x-5y+82=0. So A=12, B=-5, C=82, and (x 1,y 1)=(-1,1): d= 12(-1)-5(1)+82 √ 12^2+(-5)^2 =( -12-5+82 )/(√ 144+25 )=65/(√ 169 )=65/13=5. The distance is 5 units.

Question 34

Find the points on the x -axis, whose distances from the line x/3+y/4=1 are 4 units.

Accepted Answer

(-2, 0) and (8,0). Perpendicular distance formula with an unknown point. Any point on the x -axis has the form (a,0). Put the line in general form: x/3+y/4=1 ⇒ 4x+3y=12 ⇒ 4x+3y-12=0. Distance of (a,0) from this line: d= 4a+3(0)-12 √ 4^2+3^2 =( 4a-12 )/5. Setting d=4: 4a-12 =20 ⇒ 4a-12=20 or 4a-12=-20, a=8 or a=-2. Check: ( 4(8)-12 )/5=20/5=4 and ( 4(-2)-12 )/5=20/5=4. ✓ The points are (8,0) and (-2,0).

Question 35

Find the distance between parallel lines (i) 15 x+8 y-34=0 and 15 x+8 y+31=0 (ii) l(x+y)+p=0 and l(x+y)-r=0.

Accepted Answer

(i) 65/17 units, (ii) 1/(√ 2 ) (p+r)/l units.. Distance between parallel lines. For Ax+By+C 1=0 and Ax+By+C 2=0, d= C 1-C 2 √ A^2+B^2. (i) 15x+8y-34=0 and 15x+8y+31=0. Here A=15, B=8, C 1=-34, C 2=31: d= -34-31 √ 15^2+8^2 =65/(√ 225+64 )=65/(√ 289 )=65/17. (ii) l(x+y)+p=0 and l(x+y)-r=0, i.e. lx+ly+p=0 and lx+ly-r=0. Here A=B=l, C 1=p, C 2=-r: d= p-(-r) √ l^2+l^2 =( p+r )/( l √ 2 ). The distances are (i) 65/17 units and (ii) ( p+r )/(√ 2 l ) units.

Question 36

Find equation of the line parallel to the line 3 x-4 y+2=0 and passing through the point (-2, 3).

Accepted Answer

3 x-4 y+18=0. Family of parallel lines. Any line parallel to 3x-4y+2=0 has the same coefficients of x and y, so it is of the form 3x-4y+k=0 for some constant k (this keeps the slope 34 unchanged). It passes through (-2,3), so 3(-2)-4(3)+k=0 ⇒ -6-12+k=0 ⇒ k=18. Check: 3(-2)-4(3)+18=-6-12+18=0. ✓ The required line is 3x-4y+18=0.

Question 37

Find equation of the line perpendicular to the line x-7 y+5=0 and having x intercept 3.

Accepted Answer

y+7 x=21. Slope form, using the perpendicularity condition. Write the given line in slope-intercept form: x-7y+5=0 ⇒ y=1/7x+5/7, m 1=1/7. Two lines are perpendicular when m 1m 2=-1, so the required slope is m 2=-1/m 1=-7. An x -intercept of 3 means the line passes through (3,0). Using the point-slope form y-y 1=m(x-x 1): y-0=-7(x-3) ⇒ y=-7x+21. 7x+y-21=0

Question 38

Find angles between the lines √ 3 x+y=1 and x+√ 3 y=1.

Accepted Answer

30° and 150°. Angle between two lines. Put each line in slope form: √ 3 x+y=1 ⇒ y=-√ 3 x+1, m 1=-√ 3, x+√ 3 y=1 ⇒ y=-1/(√3)x+1/(√3), m 2=-1/(√3). If θ is an angle between the lines, then tan θ= m 2-m 1/(1+m 1m 2). Here m 1m 2=(-√3) (-1/(√3))=1, and m 2-m 1=-1/(√3)+√3=(-1+3)/(√3)=2/(√3). So tan θ= (2/√3)/(1+1) =1/(√3) ⇒ θ=30^°. The two lines cut each other in a pair of supplementary angles, so the other angle is 180^°-30^°=150^°. The angles between the lines are 30^° and 150^°.

Question 39

The line through the points (h, 3) and (4,1) intersects the line 7 x-9 y-19=0. at right angle. Find the value of h.

Accepted Answer

22/9. Perpendicularity condition m 1m 2=-1. Slope of the given line: 7x-9y-19=0 ⇒ y=7/9x-19/9, so m 1=7/9. Slope of the line through (h,3) and (4,1): m 2=1-3/4-h=-2/4-h. The two lines meet at right angles, so m 1m 2=-1: 7/9·-2/4-h=-1 ⇒ -14/(9(4-h))=-1 ⇒ 14=9(4-h)=36-9h, 9h=22 ⇒ h=22/9. Check: m 2=-2/(4-22/9)=-2/(14/9)=-9/7, and 7/9×(-9/7)=-1. ✓ h=22/9

Question 40

Prove that the line through the point (x 1, y 1) and parallel to the line A x+ B y+ C =0 is A (x-x 1)+ B (y-y 1)=0.

Accepted Answer

Equal slopes for parallel lines. Case 1: B ≠ 0. Write A x+ B y+ C =0 as y=-( A )/( B )x-( C )/( B ), so its slope is -( A )/( B ). Parallel lines have equal slopes, so the required line through (x 1,y 1) has slope -( A )/( B ) too. By the point-slope form, y-y 1=-( A )/( B )(x-x 1). Multiplying through by B (allowed, since B ≠ 0 ): B (y-y 1)=- A (x-x 1) ⇒ A (x-x 1)+ B (y-y 1)=0. Case 2: B =0 (so A ≠ 0 ). The given line is A x+ C =0, a line parallel to the y -axis; any line parallel to it is x=x 1. And with B =0 the claimed equation reduces to A (x-x 1)=0, i.e. x=x 1 — the same line. Hence in every case the line through (x 1,y 1) parallel to A x+ B y+ C =0 is A (x-x 1)+ B (y-y 1)=0.

Question 41

Two lines passing through the point (2,3) intersects each other at an angle of 60°. If slope of one line is 2, find equation of the other line.

Accepted Answer

(√ 3 +2) x+(2 √ 3 -1) y=8 √ 3 +1 or (√ 3 -2) x+(1+2 √ 3 ) y=-1+8 √ 3. Angle between two lines. Let m be the slope of the other line. With m 1=2 and the angle between the lines equal to 60^°, tan 60^°= m-2/(1+2m) ⇒ m-2/(1+2m)=±√3. Taking +√3: m-2=√3+2√3 m ⇒ m(1-2√3)=2+√3 ⇒ m=(2+√3)/(1-2√3)=-(8+5√3)/11, after multiplying numerator and denominator by 1+2√3. Taking -√3: m-2=-√3-2√3 m ⇒ m(1+2√3)=2-√3 ⇒ m=(2-√3)/(1+2√3)=(5√3-8)/11. Each line passes through (2,3), so y-3=m(x-2): For m=-(8+5√3)/11: 11(y-3)=-(8+5√3)(x-2) ⇒ (8+5√3)x+11y-(49+10√3)=0. For m=(5√3-8)/11: 11(y-3)=(5√3-8)(x-2) ⇒ (5√3-8)x-11y+(49-10√3)=0. y-3=(2+√3)/(1-2√3)(x-2) or y-3=(2-√3)/(1+2√3)(x-2); equivalently (8+5√3)x+11y-(49+10√3)=0 or (5√3-8)x-11y+(49-10√3)=0.

Question 42

Find the equation of the right bisector of the line segment joining the points (3,4) and (-1, 2).

Accepted Answer

2 x+y=5. Perpendicular bisector: midpoint plus negative reciprocal slope. The right bisector of a segment passes through its midpoint and is perpendicular to it. Midpoint of (3,4) and (-1,2): ((3+(-1))/2, (4+2)/2)=(1,3). Slope of the segment: m=2-4/-1-3=-2/-4=1/2. Slope of the bisector =-1/m=-2. Through (1,3): y-3=-2(x-1) ⇒ y-3=-2x+2. 2x+y-5=0

Question 43

Find the coordinates of the foot of perpendicular from the point (-1,3) to the line 3 x-4 y-16=0.

Accepted Answer

(68/25,-49/25). Drop a perpendicular, then solve the two lines simultaneously. The foot of the perpendicular is the point where the given line meets the perpendicular drawn to it from (-1,3). Slope of 3x-4y-16=0 is 3/4, so the perpendicular from (-1,3) has slope -4/3: y-3=-4/3(x+1) ⇒ 3y-9=-4x-4 ⇒ 4x+3y-5=0. Solve 3x-4y=16 and 4x+3y=5. Multiply the first by 3 and the second by 4 and add: 9x-12y=48, 16x+12y=20 ⇒ 25x=68 ⇒ x=68/25. Then 3y=5-4·68/25=125-272/25=-147/25, so y=-49/25. Check: 3 (68/25)-4 (-49/25)=(204+196)/25=16. ✓ The foot of the perpendicular is (68/25, -49/25).

Question 44

The perpendicular from the origin to the line y=m x+c meets it at the point (-1, 2). Find the values of m and c.

Accepted Answer

m=1/2, c=5/2. Perpendicularity plus substitution of the point. Let P (-1,2) be the foot of the perpendicular from the origin O (0,0). Slope of OP: m OP =2-0/-1-0=-2. Since OP is perpendicular to the line y=mx+c, m·(-2)=-1 ⇒ m=1/2. P (-1,2) lies on the line, so 2=1/2(-1)+c ⇒ c=2+1/2=5/2. m=1/2, c=5/2

Question 45

If p and q are the lengths of perpendiculars from the origin to the lines x cos θ-y sin θ=k cos 2 θ and x sec θ+y cosec θ=k, respectively, prove that p^2+4 q^2=k^2.

Accepted Answer

Distance of a line from the origin. The distance of A x+ B y+ C =0 from the origin is C √ A^2+ B^2. First line: x cos θ-y sin θ-k cos 2θ=0. Here √ cos^2θ+ sin^2θ =1, so p=( k cos 2θ )/1= k cos 2θ ⇒ p^2=k^2 cos^22θ. Second line: x sec θ+y cosec θ-k=0. Now sec^2θ+ cosec^2θ=1/( cos^2θ)+1/( sin^2θ)=( sin^2θ+ cos^2θ)/( sin^2θ cos^2θ)=1/( sin^2θ cos^2θ), so √ sec^2θ+ cosec^2θ =1/( sin θ cos θ ) and q=( k )/(1/ sin θ cos θ )= k sin θ cos θ =( k sin 2θ )/2, using 2 sin θ cos θ= sin 2θ. Hence 4q^2=k^2 sin^22θ. Adding, p^2+4q^2=k^2 cos^22θ+k^2 sin^22θ=k^2( cos^22θ+ sin^22θ)=k^2. Hence p^2+4q^2=k^2.

Question 46

In the triangle ABC with vertices A (2,3), B (4,-1) and C (1,2), find the equation and length of altitude from the vertex A.

Accepted Answer

y-x=1, √ 2. Altitude = perpendicular from a vertex to the opposite side. Side BC. Slope of BC through B (4,-1) and C (1,2): m BC =(2-(-1))/1-4=3/-3=-1. Equation of BC: y+1=-1(x-4) ⇒ x+y-3=0. Altitude from A. It is perpendicular to BC, so its slope is -1/-1=1. Through A (2,3): y-3=1·(x-2) ⇒ x-y+1=0. Its length is the distance from A (2,3) to the line x+y-3=0: d= 2+3-3 √ 1^2+1^2 =2/(√2)=√2. Altitude from A: x-y+1=0; its length is √ 2 units.

Question 47

If p is the length of perpendicular from the origin to the line whose intercepts on the axes are a and b, then show that 1/(p^2)=1/(a^2)+1/(b^2).

Accepted Answer

Intercept form plus the distance formula. A line with intercepts a and b on the axes is x/a+y/b=1 ⇒ bx+ay-ab=0. Its distance from the origin is p= b· 0+a· 0-ab √ a^2+b^2 = ab √ a^2+b^2. Squaring and taking reciprocals, p^2=(a^2b^2)/(a^2+b^2) ⇒ 1/(p^2)=(a^2+b^2)/(a^2b^2)=(a^2)/(a^2b^2)+(b^2)/(a^2b^2)=1/(b^2)+1/(a^2). Hence 1/(p^2)=1/(a^2)+1/(b^2).

Question 48

Find the values of k for which the line (k-3) x-(4-k^2) y+k^2-7 k+6=0 is (a) Parallel to the x -axis, (b) Parallel to the y -axis, (c) Passing through the origin.

Accepted Answer

(a) 3, (b) ± 2, (c) 6 or 1. Read off the coefficients of the general equation. For A x+ B y+ C =0: the line is parallel to the x -axis when A =0 (and B ≠0 ), parallel to the y -axis when B =0 (and A ≠0 ), and passes through the origin when C =0. Here A =k-3, B =-(4-k^2), C =k^2-7k+6. (a) Parallel to the x -axis: k-3=0 ⇒ k=3. Then B =-(4-9)=5≠0, and the line becomes 5y-6=0, i.e. y=6/5 — indeed horizontal. So k=3. (b) Parallel to the y -axis: 4-k^2=0 ⇒ k=± 2. For k=2: -x-4=0, i.e. x=-4. For k=-2: -5x+24=0, i.e. x=24/5. Both are vertical, so k=2 or k=-2. (c) Passing through the origin: k^2-7k+6=0 ⇒ (k-1)(k-6)=0 ⇒ k=1 or k=6. For k=1 the line is -2x-3y=0 and for k=6 it is 3x+32y=0; both are genuine lines through (0,0). (a) k=3 (b) k=± 2 (c) k=1 or k=6

Question 49

Find the equations of the lines, which cut-off intercepts on the axes whose sum and product are 1 and - 6, respectively.

Accepted Answer

2 x-3 y=6,-3 x+2 y=6. Intercept form, with the intercepts as roots of a quadratic. Let the intercepts be a and b, so the line is x/a+y/b=1. Given a+b=1 and ab=-6, the numbers a,b are the roots of t^2-(a+b)t+ab=0 ⇒ t^2-t-6=0 ⇒ (t-3)(t+2)=0, so a,b = 3,-2. If a=3, b=-2: x/3+y/-2=1 ⇒ 2x-3y=6. If a=-2, b=3: x/-2+y/3=1 ⇒ -3x+2y=6 ⇒ 3x-2y+6=0. 2x-3y-6=0 and 3x-2y+6=0

Question 50

What are the points on the y -axis whose distance from the line x/3+y/4=1 is 4 units.

Accepted Answer

(0,-8/3),(0, 32/3). Distance of a point from a line. Clear the fractions in x/3+y/4=1 by multiplying by 12: 4x+3y-12=0. A point on the y -axis has the form (0,k). Its distance from the line is 4(0)+3k-12 √ 4^2+3^2 =( 3k-12 )/5. Setting this equal to 4: 3k-12 =20 ⇒ 3k-12=20 or 3k-12=-20, k=32/3 or k=-8/3. The points are (0,32/3) and (0,-8/3).

Question 51

Find perpendicular distance from the origin to the line joining the points ( cos θ, sin θ) and ( cos φ, sin φ).

Accepted Answer

Cos (φ-θ)/2. Two-point form, then the distance formula. Let P ( cos θ, sin θ) and Q ( cos φ, sin φ). Using the two-point form, y- sin θ=( sin φ- sin θ)/( cos φ- cos θ) (x- cos θ). Apply the sum-to-product identities, writing A =(θ+φ)/2: sin φ- sin θ=2 cos A sin (φ-θ)/2, cos φ- cos θ=-2 sin A sin (φ-θ)/2, so the slope is - cot A (the chord is not vertical unless sin A =0, handled below). Substituting and multiplying through by sin A: (y- sin θ) sin A =- cos A (x- cos θ) ⇒ x cos A +y sin A = cos θ cos A + sin θ sin A = cos (θ- A )= cos (θ-φ)/2. Since cos^2 A + sin^2 A =1, the distance of this line from the origin is d= cos (θ-φ)/2 √ cos^2 A + sin^2 A = cos (θ-φ)/2. (If sin A =0 the chord is the vertical line x= cos A · cos (θ-φ)/2, whose distance from the origin is again cos (θ-φ)/2, so the formula still holds.) The perpendicular distance is cos (θ-φ)/2.

Question 52

Find the equation of the line parallel to y -axis and drawn through the point of intersection of the lines x-7 y+5=0 and 3 x+y=0.

Accepted Answer

x=-5/22. Find the point of intersection, then use x= constant. Solve the two lines simultaneously. From 3x+y=0 we get y=-3x. Substituting into x-7y+5=0: x-7(-3x)+5=0 ⇒ 22x+5=0 ⇒ x=-5/22, and then y=-3(-5/22)=15/22. The point of intersection is (-5/22,15/22). Every line parallel to the y -axis has the form x=c. Passing through this point forces c=-5/22. x=-5/22, i.e. 22x+5=0

Question 53

Find the equation of a line drawn perpendicular to the line x/4+y/6=1 through the point, where it meets the y -axis.

Accepted Answer

2 x-3 y+18=0. Find where the line meets the y -axis, then use the perpendicular slope. The point. On the y -axis, x=0, so 0/4+y/6=1 ⇒ y=6. The line meets the y -axis at (0,6). The slope. Multiply x/4+y/6=1 by 12: 3x+2y=12 ⇒ y=-3/2x+6, so its slope is -3/2. The perpendicular slope is m=-1/(-3/2)=2/3. The line. Through (0,6) with slope 2/3: y-6=2/3x ⇒ 3y-18=2x. 2x-3y+18=0

Question 54

Find the area of the triangle formed by the lines y-x=0, x+y=0 and x-k=0.

Accepted Answer

k^2 square units. Find the vertices, then use the coordinate area formula. Take the lines in pairs (assume k≠ 0, otherwise all three lines are concurrent at the origin and no triangle is formed). - y-x=0 and x+y=0: adding gives 2y=0, so the vertex is O (0,0). - y-x=0 and x-k=0: x=k, y=k, so the vertex is A (k,k). - x+y=0 and x-k=0: x=k, y=-k, so the vertex is B (k,-k). A and B lie on the vertical line x=k, so take AB as the base: base = k-(-k) =2 k, and the height is the horizontal distance from O (0,0) to x=k, namely k. Area =1/2× 2 k × k =k^2. The area is k^2 square units.

Question 55

Find the value of p so that the three lines 3 x+y-2=0, p x+2 y-3=0 and 2 x-y-3=0 may intersect at one point.

Accepted Answer

5. Concurrency through a computed point. Three lines pass through one point exactly when the point of intersection of two of them lies on the third. Take the two lines that do not involve p: 3x+y-2=0, 2x-y-3=0. Adding them eliminates y: 5x-5=0, so x=1, and then y=2-3x=-1. These two lines meet at (1,-1). For all three to be concurrent, (1,-1) must satisfy px+2y-3=0: p(1)+2(-1)-3=0 ⇒ p-5=0. p=5

Question 56

If three lines whose equations are y=m 1 x+c 1, y=m 2 x+c 2 and y=m 3 x+c 3 are concurrent, then show that m 1( c 2- c 3)+m 2( c 3- c 1)+m 3( c 1- c 2)=0.

Accepted Answer

Concurrency: the meeting point of two lines lies on the third. For the first two lines to meet at all we need m 1 ≠ m 2. Equating their y -values, m 1x+c 1=m 2x+c 2 ⇒ x=c 2-c 1/m 1-m 2, and then y=m 1 (c 2-c 1/m 1-m 2)+c 1=(m 1c 2-m 1c 1+c 1m 1-c 1m 2)/m 1-m 2=m 1c 2-m 2c 1/m 1-m 2. The three lines are concurrent, so this point also lies on y=m 3x+c 3: m 1c 2-m 2c 1/m 1-m 2=m 3 (c 2-c 1/m 1-m 2)+c 3. Multiplying throughout by m 1-m 2, m 1c 2-m 2c 1=m 3c 2-m 3c 1+c 3m 1-c 3m 2. Bringing everything to the left and collecting the terms in m 1, m 2, m 3, (m 1c 2-m 1c 3)+(m 2c 3-m 2c 1)+(m 3c 1-m 3c 2)=0. m 1(c 2-c 3)+m 2(c 3-c 1)+m 3(c 1-c 2)=0, as required.

Question 57

Find the equation of the lines through the point (3,2) which make an angle of 45^° with the line x-2 y=3.

Accepted Answer

3 x-y=7, x+3 y=9. Angle between two lines. If a line of slope m makes an angle θ with a line of slope m 1, then tan θ= m-m 1/(1+mm 1). Here x-2y=3 gives y=1/2x-3/2, so m 1=1/2, and θ=45^°, tan θ=1: (m- 12)/(1+m/2) =1 ⇒ 2m-1/(2+m)=± 1. Taking +1: 2m-1=2+m, so m=3. Taking -1: 2m-1=-(2+m), so 3m=-1 and m=- 13. Now use the point-slope form through (3,2). For m=3: y-2=3(x-3), i.e. 3x-y-7=0. For m=- 13: y-2=- 13(x-3), i.e. 3y-6=-x+3, i.e. x+3y-9=0. 3x-y-7=0 and x+3y-9=0

Question 58

Find the equation of the line passing through the point of intersection of the lines 4 x+7 y-3=0 and 2 x-3 y+1=0 that has equal intercepts on the axes.

Accepted Answer

13 x+13 y=6. Point of intersection, then the equal-intercept form. Solve 4x+7y-3=0 and 2x-3y+1=0. Doubling the second gives 4x-6y+2=0; subtracting it from the first, 13y-5=0 ⇒ y=5/13, x=3y-1/2=(15/13-1)/2=1/13. So the lines meet at (1/13,5/13). A line with equal intercepts a (a≠ 0) on the axes has the intercept form x/a+y/a=1, i.e. x+y=a. Since the line passes through (1/13,5/13), a=1/13+5/13=6/13. x+y=6/13, i.e. 13x+13y-6=0

Question 59

Show that the equation of the line passing through the origin and making an angle θ with the line y=m x+c is y/x=(m ± tan,)/(1 ∓ m tan,).

Accepted Answer

Angle between two lines. Any line through the origin has the form y=Mx, that is y/x=M. It makes an angle θ with y=mx+c, whose slope is m, so tan θ= M-m/(1+mM) ⇒ M-m/(1+mM)=± tan θ. Taking the + sign: M-m= tan θ+mM tan θ ⇒ M(1-m tan θ)=m+ tan θ ⇒ M=(m+ tan θ)/(1-m tan θ). Taking the - sign: M-m=- tan θ-mM tan θ ⇒ M(1+m tan θ)=m- tan θ ⇒ M=(m- tan θ)/(1+m tan θ). The two cases pair the upper signs together and the lower signs together. y/x=(m± tan θ)/(1∓ m tan θ)

Question 60

In what ratio, the line joining (-1,1) and (5,7) is divided by the line x+y=4?

Accepted Answer

1: 2. Section formula with an unknown ratio. If the point divides the join of (x 1,y 1) and (x 2,y 2) in the ratio k:1, its coordinates are ((kx 2+x 1)/(k+1),(ky 2+y 1)/(k+1)). Let x+y=4 cut the segment joining (-1,1) and (5,7) in the ratio k:1. The point of division is (5k-1/(k+1), (7k+1)/(k+1)). This point lies on x+y=4: 5k-1/(k+1)+(7k+1)/(k+1)=4 ⇒ 12k/(k+1)=4 ⇒ 12k=4k+4 ⇒ k= 12. Since k 0 the division is internal, and the point of division is (1,3), which indeed satisfies x+y=4. The line divides the segment internally in the ratio 1:2.

Question 61

Find the distance of the line 4 x+7 y+5=0 from the point (1,2) along the line 2 x-y=0.

Accepted Answer

(23 √ 5 )/18 units. Distance measured along a given line. The distance is not the perpendicular distance here — it must be measured along 2x-y=0. First note that (1,2) lies on 2x-y=0, since 2(1)-2=0. So travel along this line from (1,2) until it meets 4x+7y+5=0. Put y=2x into 4x+7y+5=0: 4x+14x+5=0 ⇒ 18x=-5 ⇒ x=-5/18, y=-5/9. So the lines meet at Q(-5/18,-5/9). The required distance is PQ=√ (1+5/18)^2+(2+5/9)^2 =√ (23/18)^2+(23/9)^2 =23/18√ 1+4. (23√ 5 )/18 units (≈ 2.86)

Question 62

Find the direction in which a straight line must be drawn through the point (-1,2) so that its point of intersection with the line x+y=4 may be at a distance of 3 units from this point.

Accepted Answer

The line is parallel to x - axis or parallel to y -axis. Parametric (distance) form of a line. A line through (x 1,y 1) inclined at θ to the x -axis has points x=x 1+r cos θ, y=y 1+r sin θ, where r is the distance of the point from (x 1,y 1). With (x 1,y 1)=(-1,2) and r=3, the point of intersection is (-1+3 cos θ, 2+3 sin θ). It lies on x+y=4: (-1+3 cos θ)+(2+3 sin θ)=4 ⇒ 3( cos θ+ sin θ)=3 ⇒ cos θ+ sin θ=1. Dividing by √ 2, cos θ cos 45^°+ sin θ sin 45^°=1/(√2) ⇒ cos (θ-45^°)= cos 45^°, so θ-45^°=± 45^°, giving θ=90^° or θ=0^°. Check: θ=0^° gives the meeting point (2,2) and θ=90^° gives (-1,5); both lie on x+y=4 and both are 3 units from (-1,2). The line must be drawn at 0^° or 90^° to the x -axis, i.e. along y=2 (parallel to the x -axis) or along x=-1 (parallel to the y -axis).

Question 63

The hypotenuse of a right angled triangle has its ends at the points (1,3) and (- 4, 1). Find an equation of the legs (perpendicular sides) of the triangle which are parallel to the axes.

Accepted Answer

x=1, y=1. or x=-4, y=3. A leg parallel to an axis fixes one coordinate of the right-angle vertex. Let the hypotenuse have ends A(1,3) and B(-4,1), and let the right angle be at C. One leg is parallel to the x -axis and the other to the y -axis, so the vertical leg through C also passes through one end and the horizontal leg through C passes through the other. That leaves exactly two positions for C: Case 1: C=(1,1). The leg AC joins (1,3) and (1,1) — a vertical line x=1. The leg BC joins (-4,1) and (1,1) — a horizontal line y=1. Case 2: C=(-4,3). The leg through B(-4,1) and C is x=-4; the leg through A(1,3) and C is y=3. In each case the two legs are perpendicular, so the angle at C is indeed a right angle. A quick sketch of A, B and the two candidate corners makes it clear why there are exactly two such triangles. The legs are x=1 and y=1, or else x=-4 and y=3.

Question 64

Find the image of the point (3,8) with respect to the line x+3 y=7 assuming the line to be a plane mirror.

Accepted Answer

(-1, - 4).. Image in a mirror line. The image Q(h,k) of P(3,8) is fixed by two conditions: PQ is perpendicular to the mirror, and the midpoint of PQ lies on the mirror. The mirror x+3y=7 has slope - 13, so PQ has slope 3: k-8/h-3=3 ⇒ k=3h-1. 1 The midpoint ((3+h)/2,(8+k)/2) lies on x+3y=7: (3+h)/2+3·(8+k)/2=7 ⇒ 3+h+24+3k=14 ⇒ h+3k=-13. 2 Substituting (1) into (2): h+3(3h-1)=-13 ⇒ 10h=-10 ⇒ h=-1, k=3(-1)-1=-4. Check: the midpoint is (1,2) and 1+3(2)=7. ✓ The image is (-1,-4).

EXERCISE 9.1 1–11 (part 1 of 7)

Exercise 1

Exercise 2

Exercise 3

Exercise 4

Exercise 5

Exercise 6

Exercise 7

Exercise 8

Exercise 9

Exercise 10

Exercise 11