# GSEB Solutions Class 11 Maths Chapter 10 Straight Lines Ex 10.1

Gujarat Board GSEB Textbook Solutions Class 11 Maths Chapter 10 Straight Lines Ex 10.1 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 11 Maths Chapter 10 Straight Lines Ex 10.1

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?
Solution:
Given points (- 4, 5), (0, 7), (5, – 5) and (-4, -2) are plotted. They are denoted by A, B, C and D respectively.
Now divide the quadrilateral into two triangles viz. ∆ABD and ∆BCD.
Area of a triangle = $$\frac{1}{2}$$ |x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)
The vertices of ∆ABD are (- 4, 5), (0, 7) and (- 4, – 2).
∴ Area of ∆ ABD = $$\frac{1}{2}$$ |- 4(7 + 2) + 0(- 2 – 5) + (- 4)(5 – 7)|
= $$\frac{1}{2}$$ |- 36 + 8| = $$\frac{28}{2}$$ = 14 ……………… (1)
The vertices of ∆ BCD are (0, 7), (5, – 5) and (- 4, – 2).
∴ Area of ∆ BCD
= $$\frac{1}{2}$$ |0(- 5 + 2) + 5(- 2 – 7) – 4(7 + 5)|
= $$\frac{1}{2}$$ |- 45 – 48| = $$\frac{93}{2}$$ ………………….. (2)
∴ Area of quadrilateral ABCD = Area of ∆ ABD + Area of ∆ BCD
= 14 + $$\frac{93}{2}$$ [From (1) and (2)]
= $$\frac{28+93}{2}$$ = $$\frac{121}{2}$$ = 60.5 sq units.

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Question 2.
The base of an equilateral triangle with side 2a lies along the y-axis such that the mid-point of the base is at the origin. Find the vertices of the triangle.
Solution:
BC is the base of ∆ ABC such that BO = OC = a.
and AB = AC = 2a
Now, AO2 = AB2 – BO2
∴ AO = $$\sqrt{4 a^{2}-a^{2}}$$ = $$\sqrt{3}$$a.
∴ The point A lying on x-axis is ($$\sqrt{3}$$a, 0).
The point B is (0, a) and the point C is (0, – a). When A lies on the left of y-axis, the vertices of triangle are (- $$\sqrt{3}$$a, 0), (0, a) and (0, – a)

Question 3.
Find the distance between P(x1, y1) and Q(x2, y2), when
(i) PQ is parallel to y-axis
(ii) PQ is parallel to x-axis.
Solution:
(i) When PQ is parallel toy-axis, every point on this line has the same abscissa. Let x2 = x1
∴ The points P and Q are P(x1, y1), Q(x1, y2)
∴ PQ = |y2 – y1|.

(ii) When PQ is parallel to x-axis. Every point on the line has the same ordinate. Let y2 = y1
∴ The coordinates of points P and Q are P(x1, y1), P(x2, y1)
∴ PQ = |x2 – x1|.

Question 4.
Find a point on the x-axis which is equidistant from the points (7, 6) and (3, 4).
Solution:
Let the point on the x-axis be (x1, 0).
The other two points A and B are A(7, 6) and B(3, 4).
W6 have: PA = PB or PA2 = PB2
or (x1 – 7)2 + 62 = (x1 – 32 + 42.
or x12 – 14x1 + 49 + 36 = x12 – 6x1 + 9 + 16
or – 14x1 + 85 = – 6x1 + 25.
or 8x1 = 60.
∴ x1 = $$\frac{60}{8}$$ = $$\frac{15}{2}$$.
∴ The point P on x-axis, equidistant from A and B is ($$\frac{15}{2}$$, 0).

Question 5.
Find the slope of a line which passes through the origin and mid-point of the line segment joining the points P(0, – 4) and B(8, 0).
Solution:
Coordinates of mid-point of two points (x1, y1) and (x2, y2) are ($$\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}$$).
The mid-point A(0, – 4) and B(8, 0) is ($$\frac{8+0}{2}$$, $$\frac{0-4}{2}$$) i.e; (4, – 2).
∴ The slope of line joining the points (x1, y1) and (x2, y2)
= $$\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$$
Here, the two points are O(0, 0) and M(4, – 2).
∴ Slope = $$\frac{-2-0}{4-0}$$ = – $$\frac{1}{2}$$.

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 traingle.
Solution:
We have three points A(4, 4), B(3, 5) and C(- 1, – 1).
∴ Slope of AB = $$\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$$

= $$\frac{5-4}{3-4}$$ = $$\frac{1}{-1}$$ = – 1.
Slope of AC = $$\frac{-1-4}{-1-4}$$ = $$\frac{-5}{-5}$$ = 1.
Product of slopes of AB and AC = 1 × (- 1) = – 1.
∴ AB ⊥ AC
Hence, ∆ ABC is a right angled triangled.

Question 7.
Find the slope of the line which makes an angle of 30° with the positive direction of y-axis (measured anticlockwise).
Solution:
The line OP makes an angle of 30° with y-axis measured anti-clockwise.
∴ OP makes an angle of 90° + 30°
= 120° with positive direction of x-axis.
∴ Slope of OP = tan 120° = tan (180° – 60°)
= – tan 60° = – $$\sqrt{3}$$.

Question 8.
Find the value of x for which the points (x, – 1), (2, 1) and (4, 5) are collinear.
Solution:
We have the points A(x, – 1), B(2, 1) and C(4, 5).
A, B and C are collinear, if the slope of AB = slope of BC.
Slope of AB = $$\frac{1+1}{2-x}$$ = $$\frac{2}{2-x}$$.
Slope of BC = $$\frac{5-1}{4-2}$$ = $$\frac{4}{2}$$ = 2.
∴ $$\frac{2}{2-x}$$ = 2
∴ 2 – x = 1 ⇒ x = 1.

Question 9.
Without using distance formula, show that the points (- 2, 1), (4, 0), (3, 3) and (- 3, 2) are the vertices of a parallelogram.
Solution:
The given points are A(- 2, – 1), B(4, 0), C(3, 3) and D(- 3, 2).
Slope of AB = $$\frac{1}{4+2}$$ = $$\frac{1}{6}$$
Slopw of DC = $$\frac{3-2}{3+3}$$ = $$\frac{1}{6}$$

∴ Slope of AB = Slope of DC
⇒ AB || DC.
Slope of AD = $$\frac{2+1}{-3+2}$$ = $$\frac{3}{-1}$$
Slope of BC = $$\frac{3-0}{3-4}$$ = $$\frac{3}{-1}$$
∴ Slope of AD = Slope of BC.
Hence, ABCD is a parallelogram.

Question 10.
Find the angle between x-axis and the line joining the points (3, – 1) and (4, – 2).
Solution:
Slope of the line joining the points P(3, – 1) and Q(4, – 2)
= $$\frac{-2+1}{4-3}$$ = $$\frac{-1}{1}$$ = – 1.
⇒ tan θ = – 1 ⇒ θ = 135°.
So, required angle is 135°.

Question 11.
The slope of a line is double of the slope of another line. If tangent of the angle between them is $$\frac{1}{3}$$, find the slopes of the lines?
Solution:
Let m1 and m2 be the slopes of the two lines.
Given that m1 = 2m2.
Let θ be the angle between the lines.
∴ tan θ = $$\frac{1}{3}$$ = $$\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}$$.
Putting m1 = 2m2, we get
∴ $$\frac{1}{3}$$ = $$\frac{2 m_{2}-m_{2}}{1+2 m_{2} \cdot m_{2}}$$ = $$\frac{m_{2}}{1+2 m_{2}^{2}}$$
or 3m2 = 1 + 2m22 or 2m22 – 3m2 + 1 = 0.
or (2m2 – 1) × (m2 – 1) = 0
∴ m2 = 1 or $$\frac{1}{2}$$
m1 = 2m2.
∴ For m2 = 1, m1 = 2.
For m2 = $$\frac{1}{2}$$, m1 = 2 × $$\frac{1}{2}$$ = 1.
∴ Slopes of these lines are 1, $$\frac{1}{2}$$ or 2, 1.

Question 12.
A line passes through the points (x1, y1) and (h, k). If the slope of the line is m, show that
k – y1 = m(h – x1).
Solution:
Slope of the line joining the points A(x1, y1) and B(h, k)
= $$\frac{k-y_{1}}{h-x_{1}}$$ = m [Given]
∴ Cross multiplying, we get
k – y1 = m(h – x1).

Question 13.
If three points (h, 0), (a, b) and (0, k) lie on a line, show that $$\frac{a}{h}$$ + $$\frac{b}{k}$$ = 1.
Solution:
The given points are A(h, 0), B(a, b) and C(0, k).
They lie on the same line.
∴ Slope of AB = Slope of BC.
∴ Slope of AB = $$\frac{b-0}{a-h}$$ = $$\frac{b}{a-h}$$.
Slope of BC = $$\frac{k-b}{0-a}$$ = $$\frac{k-b}{-a}$$.
∴ $$\frac{b}{a-h}$$ = $$\frac{k-b}{-a}$$
So, by cross multiplication, we get
– ab = (a – h)(k – b).
or – ab = ak – ab – hk + hb
or 0 = ak – hk + hb
or ak + hb = hk.
Dividing by hk, we get
$$\frac{ak}{hk}$$ + $$\frac{hb}{hk}$$ = 1
or $$\frac{a}{h}$$ + $$\frac{b}{k}$$ = 1.

Question 14.
Consider the following population and year graph. Find the slope of the line AB and using it, find what will be the population in the year 2010?

Solution:
Slope of the line joining the points A(1985, 92) and B(1995, 97)
= $$\frac{97-92}{1995-1985}$$ = $$\frac{5}{10}$$ = $$\frac{1}{2}$$.
Let p be the population in the year 2010.
∴ The point P(2010, p) lies on it ⇒ Slope of BP = Slope of AB.
∴ Slope of BP = $$\frac{p-97}{2010-1995}$$ = $$\frac{1}{2}$$.
or 2(p – 27) = 2010 – 1995 = 15.
⇒ p – 97 = $$\frac{15}{2}$$.
or p = 97 + 7.5 = 104.5 crores.