Straight Lines
Exercise 9.1 Solutions
Q1
1. Draw a quadrilateral with vertices (-4, 5), (0, 7), (5, -5), (-4, -2) and find its area.
Method
Split into two triangles ABC and ADC by joining diagonal AC.
Area Formula
Area $(\Delta) = \frac{1}{2} |x_1(y_2 – y_3) + x_2(y_3 – y_1) + x_3(y_1 – y_2)|$
Area ABC
A(-4,5), B(0,7), C(5,-5)
$= \frac{1}{2} |-4(7 – (-5)) + 0(-5 – 5) + 5(5 – 7)|$
$= \frac{1}{2} |-4(12) + 0 + 5(-2)| = \frac{1}{2} |-48 – 10| = \frac{58}{2} = 29$
$= \frac{1}{2} |-4(7 – (-5)) + 0(-5 – 5) + 5(5 – 7)|$
$= \frac{1}{2} |-4(12) + 0 + 5(-2)| = \frac{1}{2} |-48 – 10| = \frac{58}{2} = 29$
Area ADC
A(-4,5), D(-4,-2), C(5,-5)
$= \frac{1}{2} |-4(-2 – (-5)) + (-4)(-5 – 5) + 5(5 – (-2))|$
$= \frac{1}{2} |-4(3) + (-4)(-10) + 5(7)| = \frac{1}{2} |-12 + 40 + 35| = \frac{63}{2} = 31.5$
$= \frac{1}{2} |-4(-2 – (-5)) + (-4)(-5 – 5) + 5(5 – (-2))|$
$= \frac{1}{2} |-4(3) + (-4)(-10) + 5(7)| = \frac{1}{2} |-12 + 40 + 35| = \frac{63}{2} = 31.5$
Total Area
$29 + 31.5 = 60.5$ sq units.
Area = 60.5 sq. units
Q2
2. Equilateral triangle base 2a lies on y-axis, midpoint at origin. Find vertices.
Base
Midpoint is $(0,0)$. Length is $2a$. Vertices on y-axis are $(0, a)$ and $(0, -a)$.
Height
In equilateral triangle side $2a$, altitude $h = \sqrt{(2a)^2 – a^2} = \sqrt{3a^2} = a\sqrt{3}$.
3rd Vertex
Lies on x-axis at distance $h$ from origin. Can be positive or negative.
Coords
$(a\sqrt{3}, 0)$ or $(-a\sqrt{3}, 0)$.
Vertices: $(0,a), (0,-a), (\pm a\sqrt{3}, 0)$
Q3
3. Distance between $P(x_1, y_1)$ and $Q(x_2, y_2)$ when: (i) PQ || y-axis, (ii) PQ || x-axis.
(i) || y-axis
If parallel to y-axis, x-coordinates are same ($x_1 = x_2$).
Distance = $|y_2 – y_1|$.
Distance = $|y_2 – y_1|$.
(ii) || x-axis
If parallel to x-axis, y-coordinates are same ($y_1 = y_2$).
Distance = $|x_2 – x_1|$.
Distance = $|x_2 – x_1|$.
(i) $|y_2 – y_1|$, (ii) $|x_2 – x_1|$
Q4
4. Find point on x-axis equidistant from (7, 6) and (3, 4).
Let Point
$P(x, 0)$ be the point on x-axis.
Condition
$PA = PB \Rightarrow PA^2 = PB^2$.
Equation
$(x-7)^2 + (0-6)^2 = (x-3)^2 + (0-4)^2$
Expand
$x^2 – 14x + 49 + 36 = x^2 – 6x + 9 + 16$
Solve
$-14x + 85 = -6x + 25 \Rightarrow 60 = 8x \Rightarrow x = \frac{60}{8} = \frac{15}{2}$.
Point: $(\frac{15}{2}, 0)$
Q5
5. Find slope of line passing through origin and mid-point of P(0, -4) and B(8, 0).
Midpoint M
$M = (\frac{0+8}{2}, \frac{-4+0}{2}) = (4, -2)$.
Line
Passes through Origin $O(0,0)$ and $M(4, -2)$.
Slope m
$m = \frac{y_2 – y_1}{x_2 – x_1} = \frac{-2 – 0}{4 – 0} = \frac{-2}{4}$.
Slope = $-1/2$
Q6
6. Without Pythagoras, show (4, 4), (3, 5), (-1, -1) form a right angled triangle.
Slope AB
$m_{AB} = \frac{5-4}{3-4} = \frac{1}{-1} = -1$.
Slope AC
$m_{AC} = \frac{-1-4}{-1-4} = \frac{-5}{-5} = 1$.
Product
$m_{AB} \times m_{AC} = (-1) \times (1) = -1$.
Conclusion
Since product of slopes is -1, sides AB and AC are perpendicular.
It is a right angled triangle at A.
Q7
7. Find slope of line making angle of $30^\circ$ with positive y-axis (anticlockwise).
Angle
Angle with positive y-axis is $30^\circ$. Angle with positive x-axis $\theta = 90^\circ + 30^\circ = 120^\circ$.
Slope
$m = \tan \theta = \tan 120^\circ$.
Calc
$\tan(180^\circ – 60^\circ) = -\tan 60^\circ = -\sqrt{3}$.
Slope = $-\sqrt{3}$
Q8
8. Without distance formula, show (-2, -1), (4, 0), (3, 3), (-3, 2) are vertices of a parallelogram.
Slopes
$m_{AB} = \frac{0 – (-1)}{4 – (-2)} = \frac{1}{6}$
$m_{CD} = \frac{2 – 3}{-3 – 3} = \frac{-1}{-6} = \frac{1}{6}$
$m_{CD} = \frac{2 – 3}{-3 – 3} = \frac{-1}{-6} = \frac{1}{6}$
Opp Sides
$m_{AB} = m_{CD} \Rightarrow AB \parallel CD$.
Other Pair
$m_{BC} = \frac{3 – 0}{3 – 4} = -3$
$m_{DA} = \frac{-1 – 2}{-2 – (-3)} = \frac{-3}{1} = -3$
$m_{DA} = \frac{-1 – 2}{-2 – (-3)} = \frac{-3}{1} = -3$
Result
$m_{BC} = m_{DA} \Rightarrow BC \parallel DA$. Both pairs parallel.
It is a parallelogram.
Q9
9. Find angle between x-axis and line joining (3, -1) and (4, -2).
Slope m
$m = \frac{-2 – (-1)}{4 – 3} = \frac{-1}{1} = -1$.
Tan $\theta$
$\tan \theta = -1$.
Angle
$\tan \theta$ is negative in 2nd quadrant. $\theta = 180^\circ – 45^\circ = 135^\circ$.
Angle = $135^\circ$
Q10
10. Slope of one line is double the other. Angle tangent is 1/3. Find slopes.
Let
$m_1 = m, \quad m_2 = 2m$.
Formula
$\tan \theta = |\frac{m_2 – m_1}{1 + m_1 m_2}| \Rightarrow \frac{1}{3} = |\frac{2m – m}{1 + 2m^2}|$.
Case 1
$\frac{m}{1+2m^2} = \frac{1}{3} \Rightarrow 3m = 1+2m^2 \Rightarrow 2m^2 – 3m + 1 = 0$.
$(2m-1)(m-1)=0 \Rightarrow m = 1, 1/2$.
$(2m-1)(m-1)=0 \Rightarrow m = 1, 1/2$.
Case 2
$\frac{m}{1+2m^2} = -\frac{1}{3} \Rightarrow 3m = -1-2m^2 \Rightarrow 2m^2 + 3m + 1 = 0$.
$(2m+1)(m+1)=0 \Rightarrow m = -1, -1/2$.
$(2m+1)(m+1)=0 \Rightarrow m = -1, -1/2$.
Slopes: $(1, 2), (1/2, 1), (-1, -2), (-1/2, -1)$
Q11
11. A line passes through $(x_1, y_1)$ and $(h, k)$. Slope is m. Show $k – y_1 = m(h – x_1)$.
Definition
Slope $m$ of line passing through $(x_1, y_1)$ and $(x_2, y_2)$ is $\frac{y_2 – y_1}{x_2 – x_1}$.
Substitute
Here points are $(x_1, y_1)$ and $(h, k)$. So $m = \frac{k – y_1}{h – x_1}$.
Rearrange
Cross multiply: $m(h – x_1) = k – y_1$.
Hence Proved: $k – y_1 = m(h – x_1)$