Coordinate Geometry Additional Questions-10th Std Maths-Book Back Question And Answer
I. Multiple Choice Questions
Question 1.
If the three points (-3, 7), (a, 1), (-3, 2) are collinear then the value of “a” is
(1) 0
(2) -1
(3) -3
(4) 1
Answer:
(3) -3
Hint:
Since the three points are collinear
Area of a ∆ = 0
-3 + 2a – 21 – (7a – 3 – 6) = 0 ⇒ 2a – 24 – 7a + 9 = 0
– 5a – 15 = 0 ⇒ – 5(a + 3) = 0
a + 3 = 0 ⇒ a = -3
Question 2.
If A (5, 5), B (-5, 1), C (10, 7) lie in a straight line, then the area of ∆ ABC is …………….
(1) 132 sq.units
(2) 9 sq.units
(3) 25 sq.units
(4) 0
Answer:
(4) 0
Hint:
Area of the ∆le
Question 3.
In a rectangle ABCD, area of ∆ ABC is 312 sq. units. Then the area of rectangle is ……………
(1) 62 sq. units
(2) 31 sq. units
(3) 60 sq. units
(4) 30 sq. units
Answer:
(2) 31 sq. units
Hint:
In a rectangle area of ∆ ABC and area of ∆ ACD are equal.
Area of rectangle ABCD = 2 × 312 = 31 sq.units
Question 4.
If the points (k, 2k), (3k, 3k) and (3,1) are collinear, then k is ……………..
(1) 13
(2) – 13
(3) 23
(4) – 23
Answer:
(2) – 13
Hint:
Since the three points are collinear. Area of a ∆ = 0
3k2 + 3k + 6k – (6k2 + 9k + k) = 0 ⇒ 3k2 + 9k – 6k2 – 10k = 0
-3 k2 – k = 0 ⇒ -k(3k + 1) = 0
3k + 1 = 0 ⇒ 3 k = -1 ⇒ k = – 13
Question 5.
If the area of the triangle formed by the points (x, 2x), (-2, 6) and (3, 1) is 5 square units then x = ………….
(1) 2
(2) 35
(3) 3
(4) 5
Answer:
(1) 2
Hint:
Area of the triangle = 5 sq. units
6x – 2 + 6x – (-4x + 18 + x) = 10 ⇒ 12x – 2 – (-3x + 18) = 10
12x – 2 + 3x – 18 = 10
15x – 20 = 10 ⇒ 15x = 10 + 20 = 30
x = 3015 = 2
Question 6.
The slope of a line parallel to y-axis is equal to …………..
(1) 0
(2) -1
(3) 1
(4) not defined
Answer:
(4) not defined
Question 7.
In a rectangle PQRS, the slope of PQ = 56 then the slope of RS is ………..
(1) −56
(2) 65
(3) −65
(4) 56
Answer:
56
Hint:
In a rectangle opposite sides are parallel.
∴ Slope of the line RS is 56.
Question 8.
The y – intercept of the line y = 2x is ………
(1) 1
(2) 2
(3) 12
(4) 0
Answer:
(4) 0
Question 9.
The straight line given by the equation y = 5 is …………..
(1) Parallel to x – axis
(2) Parallel to y – axis
(3) Passes through the origin
(4) None of these
Answer:
(1) Parallel to x – axis
Question 10.
The x – intercept of the line 2x – 3y + 5 = 0 is ………….
(1) 52
(2) −52
(3) 25
(4) −25
Answer:
(2) −52
Hint:
2x – 3y + 5 = 0 ⇒ 2x – 3y = – 5
Question 11.
The lines 3x – 5y + 1 = 0 and 5x + ky + 2 = 0 are perpendicular if the value of k is ………..
(1) -5
(2) 3
(3) -3
(4) 5
Answer:
(2) 3
Hint:
Slope of the first line (m1) = −3−5 = 35
Slope of the second line (m2) = −5k
Since the two lines are perpendicular.
m1 × m2 = -1
35 × −5k = -1 ⇒ −3k = -1
-k = -3 ⇒ The value of k = 3
Question 12.
If x – y = 3 and x + 2y = 6 are the diameters of a circle then the centre is at the point ………..
(1) (0, 0)
(2) (1, 2)
(3) (1, -1)
(4) (4, 1)
Answer:
(4) (4, 1)
Hint:
Centre of the circle is the intersection of the two diameters.
Centre of the circle is (4, 1)
Question 13.
The line 4x + 3y – 12 = 0 meets the x-axis at the point ……….
(1) (4, 0)
(2) (3, 0)
(3) (-3, 0)
Answer:
(2) (3,0)
Hint:
4x + 3y – 12 = 0 meet the x-axis the value of y = 0
4x- 12 = 0 ⇒ 4x = 12
x = 124 = 3 ⇒ The point is (3, 0)
Question 14.
The equation of a straight line passing through the point (2, -7) and parallel to x-axis is ……………….
(1) x = 2
(2) x = -7
(3) y = -7
(4) y = 2
Answer:
(3) y = -7
Hint:
Equation of a line parallel to x-axis is y = -7
Question 15.
The equation of a straight line having slope 3 and y intercept – 4 is ………………
(1) 3x – y – 4 = 0
(2) 3x + y – 4 = 0
(3) 3x – y + 4 = 0
(4) 3x – y + 4 = 0
Answer:
(1) 3x – y – 4 = 0
Hint. The equation of a line is y = mx + c
y = 3 (x) + (-4) ⇒ y = 3x – 4
3x – y – 4 = 0
II. Answer the following questions:
Question 1.
If the points (3, – 4) (1, 6) and (- 2, 3) are the vertices of a triangle, find its area.
Answer:
Let the vertices A (3, – 4), B (1, 6) and C (- 2, 3)
Area of ∆ ABC = 12 [x1y2 + x2y3 + x3y1, – (x2y1 + x3y2 + x1y3)]
Area of a ∆ = 18 sq. units
Question 2.
If the area of the triangle formed by the points (1,2) (2,3) and (a, 4) is 8 sq. units, find a.
Answer:
Area of a triangle = 8 sq. units.
12 [x1y2 + x2y3 + x3y1 – (x2y1 + x3y2 + x1y3)] = 8.
12 [3 + 8 + 2a – (4 + 3a + 4)] = 8
11 + 2a – 8 – 3a= 16 ⇒ – a + 3 = 16
– a = 16 – 3 ⇒ a = -13
The value of a = -13
Question 3.
If the points A (2, 5), B (4, 6) and C (8, a) are collinear find the value of “a” using slope concept.
Answer:
Since the three points are collineal
Slope of a line = y2−y1x2−x1
Slope of AB = Slope of BC
6−54−2 = a−68−4 ⇒ 12 = a−64 ⇒ 2a – 12 = 4 ⇒ 2a = 16
a = 162 = 8 ⇒ The value of a = 8
Question 4.
If the points (x,y) is collinear with the points (a, 0) and (0, b) then prove that xa + yb = 1
Answer:
Let A (x, y), B (a, 0), C(0, b)
Since the three points are collinear
Slope of AB = Slope of BC
Slope of a line = y2−y1x2−x1
0−ya−x = b−00−a
−ya−x = b−a
ay = b (a – x)
ay = ba – bx
ay + bx = ab
Divided by ab
ayab + bxab = abab
yb + xa = 1 ⇒ xa + yb = 1
Question 5.
A straight line passes through (1, 2) and has the equation y – 2x – k = 0. Find k.
Answer:
The given line is y – 2x – k = 0
It passes through (1,2)
(2) -2 (1) -k = 0 ⇒ 2 – 2 – k = 0
0 – k = 0 ⇒ k = 0
The value of k = 0
Question 6.
If a line passes through the mid point of AB where A is (3, 0) and B is (5, 4) and makes an angle 60° with x – axis find its equation.
Answer:
Slope of a line = tan 60°
= 3–√
Equation of a line is y – y1 = m (x – x1)
y – 2 = 3–√ (x – 4)
y – 2 = 3–√ x – 4 3–√
3x−−√ – y + 2 – 43–√ = 0
Question 7.
Find the equation of the line through (3, 2) and perpendicular to the line joining (4, 5) and (1,2)
Answer:
Slope of a line = 2−51−4 ⇒ −3−3 = 1
Slope of the line perpendicular to it is – 1
Equation of the line joining -1 and (3, 2) is
y – y1 = m (x – x1) ⇒ y – 2 = -1(x – 3)
y – 2 = -x + 3 ⇒ x + y – 5 = 0
Question 8.
P and Q trisect the line segment joining the points (2, 1) and (5, – 8). If the point P lies on 2x – y + k = 0, then find the value of k.
Answer:
A line divides internally in the ratio 1 : 2
A line divide internally in the ratio l : m
The point P = (5+43,−8+23)
= (93,−63) = (3, -2)
The given line 2x – y + k = 0 passes through the point (3,-2)
2 (3) – (- 2) + k = 0
6 + 2 + k = 0
8 + k = 0
k = – 8
The value of k = – 8
Question 9.
The line 4x + 3y – 12 = 0 intersect the X, Y – axis at A and B respectively. Fine the area of ∆AOB.
Answer:
The equation of the line AB is 4x + 3y – 12 = 0
4x + 3y = 12
4x12 + 3y12 = 1 ⇒ x3 + y4 = 1
The point A is (3, 0) (it intersect the X – axis)
and B is (0, 4) (it intersect the Y – axis)
Area of ∆ AOB = 12 [x1y2 + x2y3 + x3y1 (x2y1 + x3y2 + x1y3)]
Question 10.
Find the equation of the line passing through (4, 5) and making equal intercept in the axes.
Answer:
Let the equal intercept on the axes be a, a.
Equation of the line is xa + ya = 1 (Given equal intercepts)
The line passes through (4, 5)
4a + 5a = 1 ⇒ 9a = 1 ⇒ a = 9
The equation of the line is x9 + y9 = 1
Multiply by 9
x + y – 9 = 0
Question 11.
Find the equation of the line passing through (2, – 1) and whose intercepts on the axes are equal in magnitude but opposite in sign.
Answer:
Let the x – intercept be “a” and y intercept be = “-a”
The equation of the line is
xa + y−a = 1 (y – intercept is – a)
xa – ya = 1
It passes through (2, -1)
2a – (−1)a = 1
2a + 1a = 1 ⇒ 3a = 1
a = 3
The equation of the line is
xa + yb = 1
x3 + y−3 = 1 ⇒ x3 – y3 = 1
x – y = 3
The equation is x – y – 3 = 0
Question 12.
The straight line cuts the coordinate axes at A and B. If the mid point of AB is (3,2) then find the equation of AB.
Answer:
Let the point A be (a, 0) and B be (0, b)
The point A (6, 0) and B (0, 4)
Equation of the line AB is
III. Answer the following questions
Question 1.
If the coordinates of two points A and B are (3, 4) and (5, – 2) respectively. Find the ‘ coordinates of any point “c”, if AC = BC and Area of triangle ABC = 10 sq. units.
Answer:
Let the coordinates C be (a, 6) then AC = BC
AC2 = BC2
(a – 3)2 + (b – 4)2 = (a – 5)2 + (b + 2)2
a2 + 9 – 6a + b2 + 16 – 8b = a2 + 25 – 10a + b2 + 4 – 4b
a2 + b2 + 25 – 6a – 86 = a2 + b2 + 29 – 10a + 4b
25 – 6a – 8b = 29 – 10a + 46
4a – 12b = 4 ⇒ a – 3b = 1 ………… (1)
Area of ∆ ABC = 10 sq. units
12 [x1y2 + x2y3 + x3y1 – (x2y1 + x3y2 + x1y3)] = 10
-6 + 5b + 4a – (20 – 2a + 3b) = 20
-6 + 5b + 4a – 20 + 2a – 3b = 20
6a + 2b – 26 = 20 ⇒ 6a + 2b = 46
Substitute the value of a = 7 in (2)
3 (7) + b = 23 ⇒ b = 23 – 21 = 2
The coordinate C is (7, 2)
Question 2.
The four vertices of a Quadrilateral are (1,2) (- 5,6) (7, – 4) and (k, – 2) taken in order. If the area of the Quadrilateral is 9 sq. units, find the value of k.
Answer:
Let A (1, 2) B (- 5, 6) C (7, – 4) and D (k, – 2)
Area of the
Quadrilateral ABCD = 12[(x1y2 + x2y3 + x3y4 + x4y1) – (x2y1 + x3y2 + x4y3 + x1y4)]
Area of the Quadrilateral ABCD = 3k – 9
Given area of a Quadrilateral is 9 sq. units.
3k – 9 = 9 ⇒ 3k = 18 ⇒ k = 183 = 6
The value of k = 6
Question 3.
Find the area of a triangles whose three sides are having the equations x + y = 2, x – y = 0 and x + 2y – 6 = 0.
Answer:
Find the three vertices of the triangles by solving their equation.
To find vertices A
Substitute the value of y = 4 in (1)
x + 4 = 2 ⇒ x = 2 – 4 = -2
The vertices A is (- 2, 4)
To find vertices B
Substitute the value of x = 1 in (1)
1 + y = 2 ⇒ y = 2 – 1 = 1
The vertices B is (1, 1)
To find vertices C
y = 63 = 2
Substitute the value y = 2 in (3)
x – 2 = 0 ⇒ x = 2
The vertices C is (2, 2)
Area of the ∆ BC = 3 sq. units
Question 4.
Verify the Median of a triangle divides into two triangles of equal areas whose vertices are A (4, – 6), B (3, – 2) and C (5, 2)
Answer:
Let D be the mid point of AC .
Mid point of AC = (5+42,2−62) = (92,-2)
Area of the triangle = 12 [x1y2 + x2y3 + x3y1 – (x2y1 + x3y2 + x1y3)]
Area of ∆ ADB = Area of ∆ BDC
A median divides the triangle of equal areas.
Question 5.
Find the area of the ∆ ABC with A (1, – 4) and the mid points of sides through A being (2,-1) and (0,-1)
Answer:
Let the coordinates of B and C are (a, b) and (c, d) respectively.
Sides through A are AB and AC
Mid point of AB = (x1+x22,y1+y22)
(2, -1) = (1+a2,−4+b2)
1+a2 = 2
1 + a = 4
a = 4 – 1
= 3
The point B is (3,2)
−4+b2 = -1
-4 + b = -2
b = -2 + 4
= 2
Mid point of AC = (1+c2,−4+d2)
(0,-1) = (1+c2,−4+d2)
1+c2 = 0
1 + c = 0
c = 0 – 1
= – 1
The point C is (-1,2)
−4+d2 = -1
– 4 + d = -2
d = – 2 + 4
= 2
Thus the coordinates of the vertices of ∆ ABC are A (1, – 4) B (3, 2) and C (- 1, 2)
Area of ∆ ABC = 12 [x1y2 + x2y3 + x3y1 – (x2y1 + x3y2 + x1y3)]
Area of ∆ ABC = 12 sq. units
Question 6.
Find the equation of the straight lines passing through (- 3, 10) whose sum of the intercepts is 8.
Answer:
Let the “x” intercept be “a” and y intercept be “b”
Sum of the intercepts = 8
a + b = 8 ⇒ b = 8 – a
Equation of a line is xa + yb = 1 ⇒ xa + y8−a = 1
It passes through (-3,10)
−3a + 108−a = 1
−3(8−a)+10aa(8−a) = 1
-24 + 3a + 10a = 8a – a2
-24 + 13a = 8a – a2
a2 + 5a – 24 = 0 ⇒ (a + 8) (a – 3) = 0
a + 8 = 0 (or) a – 3 = 0 ⇒ a = -8 (or) a = 3
The equation of a line is a
a = -8
x−8 + y8+8 = 1
x−8 + y16 = 1
-2x + y = 16
2x – y + 16 = 0
a = 3
x3 + y5 = 1
5x + 3y = 15
5x + 3y – 15 = 0
The equation of the lines are 2x – y + 16 = 0 (or) 5x + 3y – 15 = 0.
Question 7.
If (5, – 3), (- 5, 3), (6, 6) are the mid points of the sides of a triangle, find the equation of the sides.
Answer:
Since D, E, F are the mid points of ∆ ABC
EF || AB, FD || BC and DE || AC
Slope of a line = y2−y1x2−x1
Slope of EF = 6−36+5 = 311
Since EF || AB; Slope of AB = 311
Equation of AB is
y – y1 = m (x – x1)
y + 3 = 311 (x – 5)
3x – 15 = 11y + 33
3x – 11y – 15 – 33 = 0
3x – 11y – 48 = 0
Slope of DE = Slope of AC
Slope of DE = 3+3−5−5 = 6−10 = –610 = –35
Slope of AC = – 35
Equation of AC is
y – y1 = m (x – x1)
y – 6 = – 35 (x – 6) ⇒ 5y – 30 = -3x + 18
3x + 5y – 30 – 18 = 0 ⇒ 3x + 5y – 48 = 0
Slope of DF = Slope of BC
Slope of DF = 6+36−5 = 91 = 9
Slope of BC = 9
Equation of the line BC is
y – y1 = m(x – x1)
y – 3 = 9 (x + 5) ⇒ 9x + 45 = y – 3
9x – y + 45 + 3 = 0 ⇒ 9x – y + 48 = 0
Equation of the sides are
3x – 11y – 48 = 0 ; 9x – y + 48 = 0 and 3x + 5y – 48 = 0
Question 8.
Find the equation of the straight line passing through the point of intersection of the lines 5x – 8y + 23 = 0 and 7x + 6y – 71 = 0 and is perpendicular to the line joining the points (5,1) and (-2, 2)
Answer:
Substitute the value of x in (1)
5 (5) – 8y = – 23 ⇒ 25 – 8y = – 23
-8y = – 23 – 25 ⇒ -8y = – 48
y = 488 = 6
The point of intersection is (5,6)
Slope of a line = y2−y1x2−x1
Slope of the line joining the points (5,1) and (-2,2) = 2−1−2−5
= 1−7 = – 17
Slope of the perpendicular line is = 7
Equation of a line is
y – y1 = m(x – x1) ⇒ y – 6 = 7 (x – 5)
y – 6 = 7x – 35 ⇒ -7x + y – 6 + 35 = 0
7x – y – 29 = 0
Question 9.
Find the equation of the line passing through the point of intersection of 4x – y – 3 = 0 and x + y – 2 = 0 and perpendicular to 2x – 5y + 3 = 0.
Answer:
x = 55 = 1
Substitute the value of x = 1 in (2)
1 + y = 2
y = 2 – 1 = 1
The point of intersection is (1, 1)
Any line perpendicular to 2x – 5y + 3 = 0 is
5x + 2y + k = 0
It passes through (1,1)
5(1) + 2(1) + k = 0 ⇒ 5 + 2 + k = 0
7 + k = 0 ⇒ k = -7
The line is 5x + 2y – 7 = 0
Question 10.
Find the equation of the line through the point of intersection of the lines 2x + y – 5 = 0 and x + y – 3 = 0 and bisecting the line segment joining the points (3, – 2) and (- 5, 6).
Answer:
x = 2
Substitute the value of x = 2 in (2)
2 + y = 3
y = 3 – 2 = 1
The point of intersection is (2, 1)
Mid point of the line joining the points (3,-2) and (-5,6)
Mid point of the line
Equation of the line joining the points (2, 1) and (-1,2) is
x – 2 = -3 (y – 1)
x – 2 = -3y + 3
x + 3y – 5 = 0
The equation of the line is x + 3y – 5 = 0
