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Class 11 (FYJC / HSC)Mathematics & Statistics2026-27 Syllabus

Chapter 5 Straight Line Ex 5.3 Solutions

Complete Maharashtra State Board Balbharati & Yuvakbharati textbook solutions for Chapter 5 Straight Line Ex 5.3. Step-by-step solved exercises, numerical problems, and digest answers.

15 Solved Questions• 11 Diagrams• 2955 words

Maharashtra State Board 11th Maths Solutions Chapter 5 Straight Line Ex 5.3

Question 1 Maharashtra Board Solution
Write the equation of the line: i. parallel to the X-axis and at a distance of 5 units from it and above it. ii. parallel to the Y-axis and at a distance of 5 units from it and to the left of it. iii. parallel to the X-axis and at a distance of 4 units from the point (- 2,3).
Solution & Step-by-Step Answer:
i. Equation of a line parallel to X-axis is y = k. Since the line is at a distance of 5 units above X-axis, k = 5 ∴ The equation of the required line is y = 5.

ii. Equation of a line parallel to Y-axis is x = h. Since the line is at a distance of 5 units to the left of Y-axis, h = -5
∴ The equation of the required line is x = -5.
[Note: Answer given in the textbook is ‘y = -5
However, we found that ‘x = – 5’.]

iii. Equation of a line parallel to the X-axis is of the form y = k (k > 0 or k < 0).
Since the line is at a distance of 4 units from the point (- 2, 3),
k = 4 + 3 = 7 or k = 3- 4 = -1
∴ The equation of the required line is y = 1 or y = – 1.

Question 2 Maharashtra Board Solution
Obtain the equation of the line: i. parallel to the X-axis and making an intercept of 3 units on the Y-axis. ii. parallel to the Y-axis and making an intercept of 4 units on the X-axis.
Solution & Step-by-Step Answer:
i. Equation of a line parallel to X-axis with y-intercept ‘k’ isy = k. Here, y-intercept = 3 ∴ The equation of the required line is y = 3.

ii. Equation of a line parallel to Y-axis with x-intercept ‘h’ is x = h.
Here, x-intercept = 4
∴ The equation of the required line is x = 4.

Question 3 Maharashtra Board Solution
Obtain the equation of the line containing the point: i. A(2, – 3) and parallel to the Y-axis. ii. B(4, – 3) and parallel to the X-axis.
Solution & Step-by-Step Answer:
i. Equation of a line parallel to Y-axis is of the form x = h. Since the line passes through A(2, – 3), h = 2 ∴ The equation of the required line is x = 2.

ii. Equation of a line parallel to X-axis is of the formy = k.
Since the line passes through B(4, – 3), k = -3
∴ The equation of the required line is y = – 3.

Question 4 Maharashtra Board Solution
Find the equation of the line: i. passing through the points A(2, 0) and B(3,4) ii. passing through the points P(2, 1) and Q(2,-1)
Solution & Step-by-Step Answer:
i. The required line passes through the points A(2, 0) and B(3,4). Equation of the line in two point form is Here, (x1y1) = (2,0) and (x1,y2) = (3,4) ∴ The equation of the required line is ∴ ∴ ∴ y = 4(x – 2) ∴ y = 4x – 8 ∴ 4x – y – 8 = 0

ii. The required line passes through the points P(2, 1) and Q(2,-1).
Since both the given points have same
x co-ordinates i.e. 2,
the given points lie on the line x = 2.
∴ The equation of the required line is x = 2.

Question 5 Maharashtra Board Solution
Find the equation of the line: i. containing the origin and having inclination 60°. ii. passing through the origin and parallel to AB, where A is (2,4) and B is (1,7). iii. having slope 1/2 and containing the point (3, -2) iv. containing the point A(3, 5) and having slope 2/3 v. containing the point A(4, 3) and having inclination 120°. vi. passing through the origin and which bisects the portion of the line 3JC + y = 6 intercepted between the co-ordinate axes.
Solution & Step-by-Step Answer:
i. Given, Inclination of line = θ = 60° Slope of the line (m) = tan θ = tan 60° = Equation of the line having slope m and passing through origin (0, 0) is y = mx..‘. The equation of the required line is y = x

ii. Given, A (2, 4) and B (1, 7)
Slope of AB = = -3 1-2
Since the required line is parallel to line AB, slope of required line (m) = slope of AB
∴ m = – 3 and the required line passes through the origin.
Equation of the line having slope m and passing through origin (0, 0) is y = mx.
∴ The equation of the required line is y = – 3x

iii. Given, slope(m) = and the line passes through (3, – 2).
Equation of the line in slope point form is
y-y1= m(x-x1)
∴ The equation of the required line is
[y-(- 2)]=(x-3)
∴ 2(y + 2)=x – 3
∴ 2y + 4 = x – 3
∴ x – 2y – 7 = 0

iv. Given, slope(m) = and the line passes through (3, 5).
Equation of the line in slope point form is y-y1= m(x -x1)
∴ The equation of the required line is y – 5 = (x-3)
∴ 3 (y – 5) = 2 (x – 3)
∴ 3y – 15 = 2x – 6
∴ 2x – 3y + 9 = 0

v. Given, Inclination of line = θ = 120°
Slope of the line (m) = tan θ = tan 120°
= tan (90° + 30°)
= – cot 30°
= –
and the line passes through A(4, 3).
Equation of the line in slope point form is y-y1= m(x -x1)
∴ The equation of the required line is
y- 3 = –(x-4)
∴ y – 3 = – x + 4
∴ x + y – 3 -4 = 0

vi.

Given equation of the line is 3x +y = 6.
∴
This equation is of the form = 1,
where a = 2, b = 6
∴ The line 3x + y = 6 intersects the X-axis and Y-axis at A(2, 0) and B(0, 6) respectively. Required line is passing through the midpoint of AB.
∴ Midpoint of AB = ( ) = (1,3)
∴ Required line passes through (0, 0) and (1,3).
Equation of the line in two point form is

∴ The equation of the required line is


∴ y = 3x
∴ 3x – y = 0

Alternate Method:
Given equation of the line is 3x + y = 6 …(i)
Substitute y = 0 in (i) to get a point on X-axis.
∴ 3x + 0 = 6
∴ x = 2
Substitute x = 0 in (i) to get a point on Y-axis.
∴ 3(0) + 7 = 6
∴ y = 6
∴ The line 3x + y = 6 intersects the X-axis and Y-axis at A(2,0) and B(0,6) respectively.
Let M be the midpoint of AB.
M = = (1,3)
Slope of OM (m) = = 3
Equation of OM is of the formy = mx.
∴ The equation of the required line is y = 3x
∴ 3x – y = 0

Question 6 Maharashtra Board Solution
Line y = mx + c passes through the points A(2,1) and B(3,2). Determine m and c.
Solution & Step-by-Step Answer:
Given, A(2, 1) and B(3,2) Equation of the line in two point form is ∴ The equation of the required line is ∴ ∴ y – 1 = x – 2 ∴ y = x – 1 Comparing this equation with y = mx + c, we get m = 1 and c = – 1

Alternate Method:
Points A(2, 1) and B(3, 2) lie on the line y = mx + c.
∴ They must satisfy the equation.
∴ 2m + c = 1 …(i)
and 3m + c = 2 …(ii)
equation (ii) – equation (i) gives m = 1
Substituting m = 1 in (i), we get 2(1) + c = 1
∴ c = 1 – 2 = – 1

Question 7 Maharashtra Board Solution
Find the equation of the line having inclination 135° and making x-intercept 7.
Solution & Step-by-Step Answer:
Given, Inclination of line = 0 = 135° ∴ Slope of the line (m) = tan 0 = tan 135° = tan (90° + 45°) = – cot 45° = – 1 x-intercept of the required line is 7. ∴ The line passes through (7, 0). Equation of the line in slope point form is y – y1 = m(x – x1) ∴ The equation of the required line is y — 0 = – 1 (x – 7) ∴ y = -x + 7 ∴ x + y – 7 = 0
Question 8 Maharashtra Board Solution
The vertices of a triangle are A(3, 4), B(2, 0) and C(- 1, 6). Find the equations of the lines containing i. side BC ii. the median AD iii. the midpoints of sides AB and BC.
Solution & Step-by-Step Answer:
Vertices of AABC are A(3, 4), B(2, 0) and C(- 1, 6). i. Equation of the line in two point form is ∴ The equation of the side BC is ∴ – 3y = 6x – 12 ∴ 6x + 3y – 12 = 0 ∴ 2x + y – 4 = 0

ii. Let D be the midpoint of side BC.
Then, AD is the median through A.
∴ D =
The median AD passes through the points
A(3,4) and D( , 3)

∴ The equation of the median AD is


(y-4) = x – 3
∴ 5y – 20 = 2x – 6
∴ 2x – 5y + 14 = 0

iii. Let D and E be the midpoints of side AB and side BC respectively.
The equation of the line DE is

∴ -4(y-2) = 2x-5
∴ 2x + 4y – 13 = 0

Question 9 Maharashtra Board Solution
Find the x and y-intercepts of the following lines: i. ii. iii. 2x – 3y + 12 = 0
Solution & Step-by-Step Answer:
i. Given equation of the line is latex]+=1[/latex] This is of the form = 1, where x-intercept = a, y-intercept = b ∴ x-intercept = 3, y-intercept = 2

ii. Given equation of the line is = 1
∴ = 1
This is of the form = = 1,
where x-intercept = a, y-intercept = b
∴ x-intercept = and y-intercept =

iii. Given equation of the line is 2x – 3y + 12 = 0
∴ 2x – 3y = – 12
∴
∴
This is of the form = 1,
where x-intercept = a, y-intercept = b
∴ x-intercept = – 6 and y-intercept = 4

Question 10 Maharashtra Board Solution
Find equations of the line which contains the point A(l, 3) and the sum of whose intercepts on the co-ordinate axes is zero.
Solution & Step-by-Step Answer:
Case I: Line not passing through origin. Let the equation of the line be ………..(i) Since, the sum of the intercepts of the line is zero. ∴ a + b = 0 ∴ b = – a Substituting b = – a in (i), we get x – y = a...(ii) Since, the line passes through A(1, 3). ∴ 1 – 3 = a ∴ a = – 2 Substituting the value of a in (ii), equation of the required line is ∴ x – y = – 2, ∴ x – y + 2 = 0

Case II: Line passing through origin.
Slope of line passing through origin and
A(1, 3) is m = = 3
∴ Equation of the line having slope m and passing through origin (0, 0) is / = mx.
∴ The equation of the required line is y = 3x
∴ 3x – y = 0

Question 11 Maharashtra Board Solution
Find equations of the line containing the point A(3, 4) and making equal intercepts on the co-ordinate axes.
Solution & Step-by-Step Answer:
Case I: Line not passing through origin. Let the equation of the line be …………(i) This line passes through A(3, 4). ∴ ……………..(ii) Since, the required line make equal intercepts on the co-ordinate axes. ∴ a = b …(iii) Substituting the value of b in (ii), we get ∴ ∴ a = 7 ∴ b = 7 …[From (iii)] Substituting the values of a and b in (i), equation of the required line is = 1 ∴ x + y = 7

Case II: Line passing through origin.
Slope of line passing through origin and A(3,4) is m =
∴ Equation of the line having slope m and passing through origin (0, 0) is y = mx.
∴ The equation of the required line is 4
y = x
∴ 4x – 3y = 0

Question 12 Maharashtra Board Solution
Find the equations of the altitudes of the triangle whose vertices are A(2, 5), B(6, – 1 ) and C(- 4, – 3).
Solution & Step-by-Step Answer:
A(2, 5), B(6, – 1), C(- 4, – 3) are the vertices of ∆ABC. Let AD, BE and CF be the altitudes through the vertices A, B and C respectively of ∆ABC. ∴ Slope of AD = -5 …[∵AD ⊥ BC] Since altitude AD passes through (2, 5) and has slope – 5, equation of the altitude AD is y – 5 = -5 (x – 2) ∴ y – 5 = – 5x + 10 ∴ 5x +y -15 = 0 Now, slope of AC = Slope of BE = …[∵ BE ⊥ AC] Since altitude BE passes through (6,-1) and has slope , equation of the altitude BE is y-(-1) = (x – 6) ∴ 4 (y + 1) = – 3 (x – 6) ∴ 4y + 4 =-3x+ 18 ∴ 3x + 4y – 14 = 0 Also, slope of AB = ∴ Slope of CF = ….[∵ CF ⊥ AB] Since altitude CF passes through (- 4, – 3) and has slope, equation of the altitude CF is y-(-3) = [x-(-4)] ∴ 3 (y + 3) = 2 (x + 4) ∴ 3y + 9 = 2x + 8 ∴ 2x – 3y – 1 = 0

Question 13 Maharashtra Board Solution
Find the equations of perpendicular bisectors of sides of the triangle whose vertices are P(-1, 8), Q(4, – 2) and R(- 5, – 3).
Solution & Step-by-Step Answer:
Let A, B and C be the midpoints of sides PQ, QR and PR respectively of APQR. A is the midpoint of side PQ. Slope of perpendicular bisector of PQ is and it passes through (), 3). Equation of the perpendicular bisector of side PQ is y – 3 = (x – ) y – 3 = () ∴ 4(y – 3) = 2x – 3 ∴ 4y – 12 = 2x – 3 ∴ 2x – 4y + 9 = 0 B is the midpoint of side QR ∴ B = Slope of side QR = ∴ Slope of perpendicular bisector of QR is -9 and it passes through ∴ Equation of the perpendicular bisector of side QR is ∴ 2y + 5 = -18x – 9 ∴ 18x + 2y + 14 = 0 ∴ 9x + y + 7 = 0 C is the midpoint of side PR. Equation of the perpendicular bisector of PR is ∴ =-4(x + 3) ∴ 11(2y – 5) = – 8 (x + 3) ∴ 22y – 55 = – 8x – 24 ∴ 8x + 22y -31 = 0

Question 14 Maharashtra Board Solution
Find the co-ordinates of the orthocentre of the triangle whose vertices are A(2, – 2), B(l, 1) and C(-1,0).
Solution & Step-by-Step Answer:
Let O be the orthocentre of AABC. Let AM and BN be the altitudes of sides BC and AC respectively. Now, slope of BC = Slope of AM = -2,..[∵ AM ⊥ BC] Since AM passes through (2, – 2) and has slope -2, equation of the altitude AM is y – (- 2) = – 2 (x – 2) ∴ y + 2 = -2x + 4 ∴ 2x + y – 2 = 0 …(i) Also, slope of AC = ∴ Slope of BN = …[∵ BN ⊥ AC] Since BN passes through (1,1) and has slope , equation of the altitude BN is y – 1 = (x-1) ∴ 2y – 2 = 3x – 3 ∴ 3x – 2y – 1 = 0 …(ii) To find co-ordinates of orthocentre, we have to solve equations (i) and (ii). By (i) x 2 + (ii), we get 7x – 5 = 0 ∴ x = substituting x = in eq (i), we get 2() + y – 2 = 0 ∴ y = -2() + 2 ∴ y = ∴ Coordinates of orthocentre O =

Question 15 Maharashtra Board Solution
N(3, – 4) is the foot of the perpendicular drawn from the origin to line L. Find the equation of line L.
Solution & Step-by-Step Answer:
Slope of ON = Since line L ⊥ ON, slope of the line L is and it passes through point N(3, -4). Equation of the line in slope point form is y – y1 = m(x – x1) Equation of line L is y-(-4) = (x-3) ∴ 4(y + 4) = 3(x – 3) ∴ 4y + 16 = 3x – 9 ∴ 3x – 4y – 25 = 0

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