Question 1
Plot the quadrilateral with vertices (−4,5), (0,7), (5,−5), (−4,−2) and find its area.
Steps only
Use the vertices in order in the shoelace formula.
Σxᵢyᵢ₊₁=−58; Σyᵢxᵢ₊₁=63.
Area=|−58−63|/2=121/2.
Answer
60.5 square units.
CLASS 11 · MATHEMATICS · CHAPTER 9
Straight Lines
Build line equations from points, slopes and intercepts, then use them to find distances and angles. Short working is shown first; each answer also has a plain-language explanation.
78 worked answersAll 3 exercise setsAll 23 end-of-chapter questions
What you will learn
- Find slopes and angles between lines.
- Write equations in point-slope, intercept and general forms.
- Calculate point-to-line and parallel-line distances.
- Use line equations in geometric and real-life problems.
78 worked answers · short steps first, explanations when you need them
Exercise 9.1
Textbook pages 158–159
Question 2
An equilateral triangle has side 2a and a base on the y-axis whose midpoint is the origin. Find its vertices.
Steps only
The base endpoints are (0,a) and (0,−a).
The altitude is √[(2a)²−a²]=√3a.
The third vertex lies √3a units horizontally from the origin.
Answer
(0,a), (0,−a), and either (√3a,0) or (−√3a,0).
Question 3(i)
Find the distance between P(x₁,y₁) and Q(x₂,y₂) when PQ is parallel to the y-axis.
Steps only
Parallel to y-axis means x₁=x₂.
Distance=√[(x₂−x₁)²+(y₂−y₁)²]=|y₂−y₁|.
Answer
|y₂−y₁|.
Question 3(ii)
Find the distance between P(x₁,y₁) and Q(x₂,y₂) when PQ is parallel to the x-axis.
Steps only
Parallel to x-axis means y₁=y₂.
Distance=√[(x₂−x₁)²+(y₂−y₁)²]=|x₂−x₁|.
Answer
|x₂−x₁|.
Question 4
Find a point on the x-axis equidistant from (7,6) and (3,4).
Steps only
Let the point be (x,0).
(x−7)²+36=(x−3)²+16.
−14x+85=−6x+25, so x=15/2.
Answer
(15/2,0).
Question 5
Find the slope of the line through the origin and the midpoint of the segment joining (0,−4) and (8,0).
Steps only
Midpoint=((0+8)/2,(−4+0)/2)=(4,−2).
Slope from (0,0) to (4,−2) is −2/4.
Answer
−1/2.
Question 6
Without using Pythagoras, show that (4,4), (3,5), and (−1,−1) form a right triangle.
Steps only
Slopes from (4,4): to (3,5) is −1; to (−1,−1) is 1.
Their product is −1.
Answer
The angle at (4,4) is 90°.
Question 7
Find the slope of a line making an angle of 30° with the positive y-direction measured anticlockwise.
Steps only
Its angle from the positive x-axis is 90°+30°=120°.
m=tan120°=−tan60°=−√3.
Answer
−√3.
Question 8
Without using the distance formula, show that (−2,−1),(4,0),(3,3),(−3,2) are vertices of a parallelogram in that order.
Steps only
AB=(6,1)=DC and BC=(−1,3)=AD.
Answer
AB and DC are equal vectors, and BC and AD are equal vectors; hence both pairs of opposite sides are parallel and equal.
Question 9
Find the acute angle between the x-axis and the line joining (3,−1) and (4,−2).
Steps only
Slope=(−2+1)/(4−3)=−1.
The line direction has angle 135° from +x; the acute angle with the x-axis is 45°.
Answer
45°.
Question 10
One line has twice the slope of another. The tangent of the angle between them is 1/3. Find possible slopes.
Steps only
Let slopes be 2m and m.
|(2m−m)/(1+2m²)|=1/3.
3|m|=|1+2m²|; solve both signs to get m=1,1/2,−1,−1/2.
Answer
(2,1), (1,1/2), (−2,−1), or (−1,−1/2).
Question 11
A line through (x₁,y₁) and (h,k) has slope m. Show k−y₁=m(h−x₁).
Steps only
m=(k−y₁)/(h−x₁), when h≠x₁.
Multiply by h−x₁.
Answer
The required relation follows from the slope formula.
Exercise 9.2
Textbook pages 161–164
Question 1
Write equations for the x-axis and y-axis.
Steps only
On the x-axis, every point has y-coordinate 0.
On the y-axis, every point has x-coordinate 0.
Answer
x-axis: y=0. y-axis: x=0.
Question 2
Find the line through (−4,3) with slope 1/2.
Steps only
y−3=(1/2)(x+4).
Multiply by 2 and rearrange.
Answer
x−2y+10=0.
Question 3
Find the line through (0,0) with slope m.
Steps only
Use y−y₁=m(x−x₁).
Set x₁=y₁=0.
Answer
y=mx.
Question 4
Find the equation of the line through (3/2,2), inclined at 75° to the x-axis.
Steps only
Slope=tan75°=2+√3.
Use point-slope form through (3/2,2).
Answer
y−2=(2+√3)(x−3/2).
Question 5
Find the line with x-intercept −3 and slope −2.
Steps only
The line passes through (−3,0).
y−0=−2(x+3).
Answer
y=−2x−6.
Question 6
Find the line with y-intercept 2 and inclination 30°.
Steps only
Slope=tan30°=1/√3.
Use point (0,2): y−2=x/√3.
Answer
y=x/√3+2.
Question 7
Find the line through (−1,1) and (2,−4).
Steps only
Slope=(−4−1)/(2+1)=−5/3.
y−1=−(5/3)(x+1).
Rearrange to 5x+3y+2=0.
Answer
5x+3y+2=0.
Question 8
For triangle P(2,1), Q(−2,3), R(4,5), find the median through R.
Steps only
Midpoint of PQ=(0,2).
Slope from R(4,5) to (0,2) is 3/4.
y−5=(3/4)(x−4).
Answer
3x−4y+8=0.
Question 9
Find the line through (−3,5) perpendicular to the line through (2,5) and (−3,6).
Steps only
Given slope=(6−5)/(−3−2)=−1/5.
Perpendicular slope=5.
y−5=5(x+3).
Answer
5x−y+20=0.
Question 10
A perpendicular to the segment joining (1,0),(2,3) divides it internally in ratio 1:n. Find its equation.
Steps only
Division point=((n+2)/(n+1),3/(n+1)).
Segment slope=3, so perpendicular slope=−1/3.
Use point-slope form and clear denominators.
Answer
(n+1)x+3(n+1)y−(n+11)=0.
Question 11
Find the line cutting equal intercepts on the axes and passing through (2,3).
Steps only
Equal-intercept form is x/a+y/a=1, or x+y=a.
Substitute (2,3): a=5.
Answer
x+y=5.
Question 12
A line through (2,2) cuts intercepts on the axes whose sum is 9. Find its equation(s).
Steps only
Let intercepts be a,b: a+b=9 and 2/a+2/b=1.
Thus ab=18.
a,b are 3 and 6 in either order.
Answer
2x+y=6 or x+2y=6.
Question 13
Find the line through (0,2) at angle 2π/3 to +x, and a parallel line crossing the y-axis 2 units below the origin.
Steps only
Slope=tan(2π/3)=−√3.
Use y-intercepts 2 and −2.
Answer
y=−√3x+2 and y=−√3x−2.
Question 14
The perpendicular from the origin meets a line at (−2,9). Find the line equation.
Steps only
The perpendicular from origin to (−2,9) has direction vector (−2,9).
This is a normal vector to the line.
−2(x+2)+9(y−9)=0.
Answer
2x−9y+85=0.
Question 15
A copper rod has length 124.942 cm at 20°C and 125.134 cm at 110°C. Assuming a linear relation, express L in terms of C.
Steps only
Slope=(125.134−124.942)/(110−20)=0.192/90.
Slope=0.0021333….
Intercept=124.942−20(slope)=124.899333….
Answer
L=124.899333…+0.0021333…C cm.
Question 16
Demand is 980 L/week at ₹14/L and 1220 L/week at ₹16/L. If price and demand are linearly related, find demand at ₹17/L.
Steps only
Demand rises 240 L when price rises ₹2.
Rate=120 L per rupee.
At ₹17: 1220+120=1340.
Answer
1,340 L per week.
Question 17
P(a,b) is the midpoint of a segment between the coordinate axes. Show that the line equation is x/a+y/b=2.
Steps only
If endpoints are (u,0),(0,v), their midpoint is (u/2,v/2)=(a,b).
Thus u=2a, v=2b.
Intercept form: x/(2a)+y/(2b)=1.
Answer
x/a+y/b=2.
Question 18
R(h,k) divides a segment between the axes in ratio 1:2. Find the line equation.
Steps only
Let intercept endpoints be (a,0),(0,b).
Section formula gives h=2a/3 and k=b/3.
a=3h/2,b=3k; substitute into x/a+y/b=1.
Answer
2kx+hy=3hk.
Question 19
Use a line equation to show (3,0),(−2,−2),(8,2) are collinear.
Steps only
Line through (3,0) and (−2,−2) has slope 2/5.
Its equation is 2x−5y−6=0.
Substitute (8,2): 16−10−6=0.
Answer
All three satisfy 2x−5y−6=0.
Exercise 9.3
Textbook pages 165–168
Question 1(i)
Write x+7y=0 in slope-intercept form and give slope and y-intercept.
Steps only
7y=−x.
Divide by 7.
Answer
y=−x/7; slope −1/7; y-intercept 0.
Question 1(ii)
Write 6x+3y−5=0 in slope-intercept form and give slope and y-intercept.
Steps only
3y=−6x+5.
Divide by 3.
Answer
y=−2x+5/3; slope −2; y-intercept 5/3.
Question 1(iii)
Write y=0 in slope-intercept form and give slope and y-intercept.
Steps only
Already y=0x+0.
Answer
Slope 0; y-intercept 0.
Question 2(i)
Write 3x+2y−12=0 in intercept form and find intercepts.
Steps only
3x+2y=12.
Divide by 12.
Answer
x/4+y/6=1; intercepts 4 and 6.
Question 2(ii)
Write 4x−3y=6 in intercept form and find intercepts.
Steps only
Set y=0: x=3/2.
Set x=0: y=−2.
Answer
x/(3/2)+y/(−2)=1; intercepts 3/2 and −2.
Question 2(iii)
For 3y+2=0, identify its intercepts.
Steps only
y=−2/3 for every x.
It never reaches y=0.
Answer
y-intercept −2/3; no x-intercept.
Question 3
Find the distance from (−1,1) to 12(x+6)=5(y−2).
Steps only
Write 12x−5y+82=0.
Distance=|12(−1)−5(1)+82|/√(12²+5²)=65/13.
Answer
5 units.
Question 4
Find points on the x-axis whose distance from x/3+y/4=1 is 4.
Steps only
Line is 4x+3y−12=0.
For (x,0): |4x−12|/5=4.
Solve |4x−12|=20.
Answer
(−2,0) and (8,0).
Question 5(i)
Find the distance between 15x+8y−34=0 and 15x+8y+31=0.
Steps only
Coefficients match; distance=|31−(−34)|/√(15²+8²).
=65/17.
Answer
65/17 units.
Question 5(ii)
Find the distance between l(x+y)+p=0 and l(x+y)−r=0.
Steps only
Write coefficients A=B=l.
Difference in constants is |p−(−r)|=|p+r|.
√(A²+B²)=|l|√2.
Answer
|p+r|/(|l|√2), for l≠0.
Question 6
Find the line parallel to 3x−4y+2=0 through (−2,3).
Steps only
Parallel line has form 3x−4y+C=0.
Substitute (−2,3): −6−12+C=0, so C=18.
Answer
3x−4y+18=0.
Question 7
Find the line perpendicular to x−7y+5=0 with x-intercept 3.
Steps only
Given slope=1/7.
Perpendicular slope=−7 and point=(3,0).
y=−7(x−3).
Answer
7x+y−21=0.
Question 8
Find the angle between 13x+y=0 and x+13y=0.
Steps only
Slopes are −13 and −1/13.
tanθ=|m₂−m₁|/|1+m₁m₂|.
=|(−1/13)+13|/2=84/13.
Answer
tanθ=84/13; θ=tan⁻¹(84/13).
Question 9
The line through (h,3),(4,1) is perpendicular to 7x−9y−19=0. Find h.
Steps only
Given line slope=7/9, so perpendicular slope=−9/7.
−2/(4−h)=−9/7.
14=36−9h, so h=22/9.
Answer
h=22/9.
Question 10
Prove that the line through (x₁,y₁) parallel to Ax+By+C=0 has equation A(x−x₁)+B(y−y₁)=0.
Steps only
The original line has normal vector (A,B).
A parallel line has the same normal vector.
Use the point-normal equation through (x₁,y₁).
Answer
The equation is proved.
Question 11
Two lines through (2,3) meet at 60°. If one slope is 2, find the other slope(s).
Steps only
Use |(m−2)/(1+2m)|=tan60°=√3.
Solve m−2=±√3(1+2m).
Answer
m=(2+√3)/(1−2√3) or m=(2−√3)/(1+2√3).
Question 12
Find the perpendicular bisector of the segment joining (3,4) and (−1,2).
Steps only
Midpoint=(1,3).
Segment slope=1/2; perpendicular slope=−2.
y−3=−2(x−1).
Answer
2x+y−5=0.
Question 13
Find the foot of the perpendicular from (−1,3) to 3x−4y−16=0.
Steps only
D=3(−1)−4(3)−16=−31; A²+B²=25.
Foot=(x₀−AD/25,y₀−BD/25).
=(68/25,−49/25).
Answer
(68/25,−49/25).
Question 14
The perpendicular from the origin to y=mx+c meets it at (−1,2). Find m,c.
Steps only
The perpendicular has slope 2/(−1)=−2.
Thus the given line slope m=1/2.
Use (−1,2): 2=−1/2+c.
Answer
m=1/2, c=5/2.
Question 15
If p and q are the distances from the origin to x cosθ−y sinθ=k cos2θ and x secθ+y cosecθ=k, prove p²+4q²=k².
Steps only
First line has unit normal, so p=|k cos2θ|.
For the second, q=|k sinθ cosθ|=|k sin2θ|/2.
p²+4q²=k²(cos²2θ+sin²2θ)=k².
Answer
p²+4q²=k².
Question 16
In triangle A(2,3), B(4,−1), C(1,2), find the equation and length of the altitude from A.
Steps only
Slope BC=(2+1)/(1−4)=−1, so altitude slope=1.
Through A: y−3=x−2, or y=x+1.
BC: x+y−3=0; distance from A is |2+3−3|/√2=√2.
Answer
Equation y=x+1; length √2.
Question 17
If p is the perpendicular distance from origin to a line with intercepts a,b, prove 1/p²=1/a²+1/b².
Steps only
Line: x/a+y/b=1, or bx+ay−ab=0.
p²=a²b²/(a²+b²).
Take reciprocals.
Answer
The identity is proved.
Miscellaneous Exercise on Chapter 9
Textbook pages 171–174
Question 1(a)
For which k is (k−3)x−(4−k²)y+k²−7k+6=0 parallel to the x-axis?
Steps only
A horizontal line has x-coefficient 0.
k−3=0 gives k=3.
Answer
k=3.
Question 1(b)
For which k is (k−3)x−(4−k²)y+k²−7k+6=0 parallel to the y-axis?
Steps only
A vertical line has y-coefficient 0.
4−k²=0, so k=±2.
Answer
k=2 or k=−2.
Question 1(c)
For which k does (k−3)x−(4−k²)y+k²−7k+6=0 pass through the origin?
Steps only
At the origin the constant term must be zero.
k²−7k+6=(k−1)(k−6).
Answer
k=1 or k=6.
Question 2
Find lines whose intercepts have sum 1 and product −6.
Steps only
Let intercepts be a,b. Then a+b=1, ab=−6.
The roots are a=3,b=−2 or a=−2,b=3.
Use x/a+y/b=1.
Answer
2x−3y=6 or 3x−2y=−6.
Question 3
Find points on the y-axis at distance 4 from x/3+y/4=1.
Steps only
Line is 4x+3y−12=0.
For (0,y), |3y−12|/5=4.
Solve 3y−12=±20.
Answer
(0,32/3) and (0,−8/3).
Question 4
Find the perpendicular distance from the origin to the chord through (cosθ,sinθ) and (cosφ,sinφ) on the unit circle.
Steps only
The chord equation is x cos((θ+φ)/2)+y sin((θ+φ)/2)=cos((θ−φ)/2).
The normal vector has length 1.
Distance from origin is the absolute constant term.
Answer
|cos((θ−φ)/2)|.
Question 5
Find the line parallel to the y-axis through the intersection of x−7y+5=0 and 3x+y=0.
Steps only
From 3x+y=0, y=−3x.
Substitute: 22x+5=0, so x=−5/22.
A parallel-to-y-axis line keeps x constant.
Answer
x=−5/22.
Question 6
Find a line perpendicular to x/4+y/6=1 through the point where it meets the y-axis.
Steps only
The y-intercept is (0,6).
Original slope=−3/2; perpendicular slope=2/3.
y−6=(2/3)x.
Answer
2x−3y+18=0.
Question 7
Find the area enclosed by y−x=0, x+y=0, and x−k=0.
Steps only
The vertices are (0,0),(k,k),(k,−k).
Vertical base length=2|k|; horizontal height=|k|.
Area=(1/2)(2|k|)(|k|)=k².
Answer
k² square units.
Question 8
Find p so 3x+y−2=0, px+2y−3=0, and 2x−y−3=0 are concurrent.
Steps only
First and third lines meet at (1,−1).
Substitute in the second: p−2−3=0.
Answer
p=5.
Question 9
Prove that concurrency of y=m₁x+c₁, y=m₂x+c₂, y=m₃x+c₃ implies m₁(c₂−c₃)+m₂(c₃−c₁)+m₃(c₁−c₂)=0.
Steps only
At a common point (x,y), cᵢ=y−mᵢx.
Substitute these expressions into the left side.
All y-terms and x-terms cancel.
Answer
The condition is proved.
Question 10
Find lines through (3,2) making 45° with x−2y=3.
Steps only
Given line slope=1/2.
Solve |(m−1/2)/(1+m/2)|=1.
The slopes are 3 and −1/3; use point-slope form.
Answer
3x−y−7=0 or x+3y−9=0.
Question 11
Find the equal-intercept line through the intersection of 4x+7y−3=0 and 2x−3y+1=0.
Steps only
Solving gives intersection (1/13,5/13).
An equal-intercept line is x+y=c.
Substitute the point: c=6/13.
Answer
x+y=6/13.
Question 12
Show that a line through the origin making angle θ with y=mx+c has slope (m±tanθ)/(1∓m tanθ).
Steps only
The line y=mx+c has inclination α where tanα=m.
The new line has inclination α±θ.
tan(α±θ)=(m±tanθ)/(1∓m tanθ).
Answer
The stated slope formula follows from the tangent addition/subtraction identities.
Question 13
In what ratio does the line joining (−1,1) and (5,7) meet x+y=4?
Steps only
A point at fraction t from (−1,1) is (−1+6t,1+6t).
Set coordinate sum 4: 12t=4, so t=1/3.
Answer
1:2 internally.
Question 14
Find the distance from (1,2) to 4x+7y+5=0 measured along the line 2x−y=0.
Steps only
The measuring line through (1,2) is y=2x.
Its intersection with the target line is (−5/18,−5/9).
Use the distance formula between that point and (1,2).
Answer
23√5/18 units.
Question 15
Find lines through (−1,2) whose intersection with x+y=4 is 3 units away.
Steps only
Let intersection be (x,4−x).
(x+1)²+(2−x)²=9 gives x²−x−2=0.
x=2 or −1, giving (2,2) or (−1,5).
Answer
y=2 or x=−1.
Question 16
A right triangle has hypotenuse endpoints (1,3),(−4,1), and its legs are parallel to the axes. Give the possible leg equations.
Steps only
The right-angle vertex must combine one endpoint’s x-coordinate with the other endpoint’s y-coordinate.
Possible corners: (1,1) or (−4,3).
Answer
Either x=1 and y=1, or y=3 and x=−4.
Question 17
Find the reflection of (3,8) across x+3y=7.
Steps only
The perpendicular foot on the line is (1,2).
Reflect by doubling the foot and subtracting the original point.
(2(1)−3,2(2)−8)=(−1,−4).
Answer
(−1,−4).
Question 18
The lines y=3x+1 and 2y=x+3 are equally inclined to y=mx+4. Find m.
Steps only
Set |(3−m)/(1+3m)|=|(1/2−m)/(1+m/2)|.
The consistent sign case reduces to 7m²−2m−7=0.
Solve the quadratic.
Answer
m=(1±5√2)/7.
Question 19
A point has the sum of its perpendicular distances from x+y−5=0 and 3x−2y+7=0 equal to 10. Describe its locus.
Steps only
The point-to-line formula gives distances |x+y−5|/√2 and |3x−2y+7|/√13.
Set their sum equal to 10 to get the locus equation.
In the coordinates u=x+y−5 and v=3x−2y+7, this becomes |u|/√2+|v|/√13=10, a diamond. Transforming its four corners back gives the listed parallelogram vertices.
Answer
The boundary of a parallelogram: |x+y−5|/√2 + |3x−2y+7|/√13 = 10. Its vertices are ((3±20√2)/5,(22±30√2)/5) and ((3±10√13)/5,(22∓10√13)/5), with each sign paired.
Question 20
Find the line equidistant from 9x+6y−7=0 and 3x+2y+6=0.
Steps only
Put both in matching scale: 3x+2y−7/3=0 and 3x+2y+6=0.
Midpoint constant=(−7/3+6)/2=11/6.
Multiply by 6.
Answer
18x+12y+11=0.
Question 21
A ray from (1,2) reflects at the x-axis and then passes through (5,3). Find reflection point A.
Steps only
Reflect (5,3) across x-axis to (5,−3).
Line from (1,2) to (5,−3) meets y=0 at parameter t=2/5.
x=1+4(2/5)=13/5.
Answer
A=(13/5,0).
Question 22
For the textbook’s points ±(√(a²−b²),0) and line (x/a)cosθ+(y/b)sinθ=1, prove the product of perpendicular distances is b².
Steps only
Let r=√(a²−b²). The line has normal length N=√(cos²θ/a²+sin²θ/b²).
Distances are |r cosθ/a−1|/N and |−r cosθ/a−1|/N.
Their product=[1−r²cos²θ/a²]/N²=b².
Answer
The product is b².
Question 23
At the intersection of 2x−3y+4=0 and 3x+4y−5=0, find the path to 6x−7y+8=0 that takes the least time at constant speed.
Steps only
Starting point is (−1/17,22/17).
Shortest path to a line is perpendicular to it.
Target slope=6/7, so path slope=−7/6; use point-slope form.
Answer
119x+102y−125=0.