Class 11 Maths Exercise 9.1 Solutions: Slope, Distance, and Quadrilateral Area (Straight Lines)
This exercise covers foundational concepts in coordinate geometry: distance, slope, area of a quadrilateral, and properties of geometric figures.
1. Quadrilateral Area
The vertices of the quadrilateral are $A(-4, 5)$, $B(0, 7)$, $C(5, -5)$, and $D(-4, -2)$.
We can find the area of the quadrilateral $ABCD$ by dividing it into two triangles, for example, $\triangle ABC$ and $\triangle ADC$.
The Area of a Triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is:
$$\text{Area} = \frac{1}{2} |x_1(y_2 – y_3) + x_2(y_3 – y_1) + x_3(y_1 – y_2)|$$
Area of $\triangle ABC$: $A(-4, 5), B(0, 7), C(5, -5)$
Let $(x_1, y_1) = (-4, 5)$, $(x_2, y_2) = (0, 7)$, $(x_3, y_3) = (5, -5)$.
$$\text{Area}(ABC) = \frac{1}{2} |(-4)(7 – (-5)) + (0)(-5 – 5) + (5)(5 – 7)|$$
$$\text{Area}(ABC) = \frac{1}{2} |(-4)(12) + 0 + (5)(-2)|$$
$$\text{Area}(ABC) = \frac{1}{2} |-48 – 10| = \frac{1}{2} |-58| = 29 \text{ sq. units}$$
Area of $\triangle ADC$: $A(-4, 5), D(-4, -2), C(5, -5)$
Let $(x_1, y_1) = (-4, 5)$, $(x_2, y_2) = (-4, -2)$, $(x_3, y_3) = (5, -5)$.
$$\text{Area}(ADC) = \frac{1}{2} |(-4)(-2 – (-5)) + (-4)(-5 – 5) + (5)(5 – (-2))|$$
$$\text{Area}(ADC) = \frac{1}{2} |(-4)(3) + (-4)(-10) + (5)(7)|$$
$$\text{Area}(ADC) = \frac{1}{2} |-12 + 40 + 35|$$
$$\text{Area}(ADC) = \frac{1}{2} |63| = 31.5 \text{ sq. units}$$
Area of Quadrilateral $ABCD$
$$\text{Area}(ABCD) = \text{Area}(ABC) + \text{Area}(ADC)$$
$$\text{Area}(ABCD) = 29 + 31.5 = \mathbf{60.5 \text{ sq. units}}$$
2. Vertices of an Equilateral Triangle
Given:
- Side length $= 2a$.
- Base lies along the $y$-axis.
- Mid-point of the base is at the origin $(0, 0)$.
Since the base lies on the $y$-axis and its mid-point is $(0, 0)$, the base vertices must be equidistant from the origin along the $y$-axis.
Side length is $2a$, so half the side is $a$.
The base vertices are $B_1 = \mathbf{(0, a)}$ and $B_2 = \mathbf{(0, -a)}$.
Let the third vertex be $V(x, y)$. Since the triangle is equilateral, the distance from $V$ to $B_1$ and $B_2$ must be $2a$, and the distance from $V$ to the $y$-axis (the $x$-coordinate) must be such that the Pythagorean theorem holds for the right triangle formed by the height.
The height ($h$) of an equilateral triangle with side $s$ is $h = \frac{\sqrt{3}}{2} s$.
Here, $s = 2a$, so $h = \frac{\sqrt{3}}{2} (2a) = a\sqrt{3}$.
Since the base is on the $y$-axis, the third vertex $V$ must lie on the $x$-axis or a line parallel to the $x$-axis, $a\sqrt{3}$ units away from the $y$-axis.
The height is the absolute value of the $x$-coordinate of $V$, so $|x| = a\sqrt{3}$.
$$x = \pm a\sqrt{3}$$
The two possible sets of vertices for the triangle are:
$$ \mathbf{(0, a), (0, -a), \text{ and } (a\sqrt{3}, 0)}$$
$$\text{OR}$$
$$ \mathbf{(0, a), (0, -a), \text{ and } (-a\sqrt{3}, 0)}$$
(Verification using distance formula for $V(a\sqrt{3}, 0)$ and $B_1(0, a)$):
$$d^2 = (a\sqrt{3} – 0)^2 + (0 – a)^2 = 3a^2 + a^2 = 4a^2$$
$$d = 2a \text{ (Side length). Correct.}$$
3. Distance Between Two Points
Find the distance between $P(x_1, y_1)$ and $Q(x_2, y_2)$.
The Distance Formula is $d = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}$.
(i) $PQ$ is parallel to the $y$-axis
If $PQ$ is parallel to the $y$-axis, the $x$-coordinates must be the same: $x_1 = x_2$.
$$d = \sqrt{(x_1 – x_1)^2 + (y_2 – y_1)^2}$$
$$d = \sqrt{0 + (y_2 – y_1)^2} = |y_2 – y_1|$$
The distance is $\mathbf{|y_2 – y_1|}$.
(ii) $PQ$ is parallel to the $x$-axis
If $PQ$ is parallel to the $x$-axis, the $y$-coordinates must be the same: $y_1 = y_2$.
$$d = \sqrt{(x_2 – x_1)^2 + (y_1 – y_1)^2}$$
$$d = \sqrt{(x_2 – x_1)^2 + 0} = |x_2 – x_1|$$
The distance is $\mathbf{|x_2 – x_1|}$.
4. Equidistant Point on the $x$-axis
Find a point $P(x, 0)$ on the $x$-axis that is equidistant from $A(7, 6)$ and $B(3, 4)$.
We require $PA = PB$, or $PA^2 = PB^2$.
$$PA^2 = (x – 7)^2 + (0 – 6)^2 = (x – 7)^2 + 36$$
$$PB^2 = (x – 3)^2 + (0 – 4)^2 = (x – 3)^2 + 16$$
Set $PA^2 = PB^2$:
$$(x – 7)^2 + 36 = (x – 3)^2 + 16$$
Expand the squares:
$$(x^2 – 14x + 49) + 36 = (x^2 – 6x + 9) + 16$$
$$x^2 – 14x + 85 = x^2 – 6x + 25$$
Subtract $x^2$ from both sides:
$$-14x + 85 = -6x + 25$$
$$-14x + 6x = 25 – 85$$
$$-8x = -60$$
$$x = \frac{60}{8} = \frac{15}{2}$$
The point on the $x$-axis is $\mathbf{\left(\frac{15}{2}, 0\right)}$ or $\mathbf{(7.5, 0)}$.
5. Slope of a Line Through Origin and Mid-point
Find the slope of a line passing through the origin $O(0, 0)$ and the mid-point $M$ of the segment joining $P(0, -4)$ and $B(8, 0)$.
1. Find the Mid-point $M$
The mid-point formula is $M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$.
$$M = \left(\frac{0 + 8}{2}, \frac{-4 + 0}{2}\right) = \left(\frac{8}{2}, \frac{-4}{2}\right) = \mathbf{(4, -2)}$$
2. Find the Slope of Line $OM$
The slope $m$ of a line passing through $(x_1, y_1)$ and $(x_2, y_2)$ is $m = \frac{y_2 – y_1}{x_2 – x_1}$.
Using $O(0, 0)$ and $M(4, -2)$:
$$m = \frac{-2 – 0}{4 – 0} = \frac{-2}{4} = \mathbf{-\frac{1}{2}}$$
6. Right Angled Triangle (Using Slopes)
Show that the points $A(4, 4)$, $B(3, 5)$, and $C(-1, -1)$ are the vertices of a right angled triangle without using the distance formula.
A triangle is a right angled triangle if two of its sides are perpendicular. Two lines with slopes $m_1$ and $m_2$ are perpendicular if and only if $m_1 m_2 = -1$.
Calculate the slopes of the three sides ($m_{AB}, m_{BC}, m_{AC}$):
Slope of $AB$, $m_{AB}$: $A(4, 4), B(3, 5)$
$$m_{AB} = \frac{5 – 4}{3 – 4} = \frac{1}{-1} = \mathbf{-1}$$
Slope of $BC$, $m_{BC}$: $B(3, 5), C(-1, -1)$
$$m_{BC} = \frac{-1 – 5}{-1 – 3} = \frac{-6}{-4} = \mathbf{\frac{3}{2}}$$
Slope of $AC$, $m_{AC}$: $A(4, 4), C(-1, -1)$
$$m_{AC} = \frac{-1 – 4}{-1 – 4} = \frac{-5}{-5} = \mathbf{1}$$
Check for perpendicularity:
- $m_{AB} \cdot m_{BC} = (-1) \cdot (\frac{3}{2}) = -\frac{3}{2} \ne -1$
- $m_{BC} \cdot m_{AC} = (\frac{3}{2}) \cdot (1) = \frac{3}{2} \ne -1$
- $m_{AB} \cdot m_{AC} = (-1) \cdot (1) = -1$
Since $m_{AB} \cdot m_{AC} = -1$, the side $AB$ is perpendicular to the side $AC$.
Therefore, the triangle $ABC$ has a right angle at $A$, and the points are the vertices of a right angled triangle.
7. Slope with Angle from $y$-axis
Find the slope of the line which makes an angle of $\mathbf{30^\circ}$ with the positive direction of the $y$-axis measured anticlockwise.
The slope $m$ of a line is defined as $m = \tan \theta$, where $\theta$ is the angle the line makes with the positive direction of the $x$-axis measured anticlockwise.
The line forms a $30^\circ$ angle with the positive $y$-axis. The $x$-axis and $y$-axis are perpendicular, meaning the angle between them is $90^\circ$.
The angle $\theta$ with the positive $x$-axis is:
$$\theta = 90^\circ + 30^\circ = 120^\circ$$
The slope is:
$$m = \tan(120^\circ)$$
$$m = \tan(180^\circ – 60^\circ) = -\tan(60^\circ) = \mathbf{-\sqrt{3}}$$
8. Parallelogram Vertices (Using Slopes)
Show that points $A(-2, -1)$, $B(4, 0)$, $C(3, 3)$, and $D(-3, 2)$ are the vertices of a parallelogram without using the distance formula.
A quadrilateral is a parallelogram if opposite sides are parallel. Lines are parallel if and only if they have the same slope. We check the slopes of $AB$ and $CD$ (opposite) and $BC$ and $AD$ (opposite).
Slope of $AB$, $m_{AB}$: $A(-2, -1), B(4, 0)$
$$m_{AB} = \frac{0 – (-1)}{4 – (-2)} = \frac{1}{6}$$
Slope of $CD$, $m_{CD}$: $C(3, 3), D(-3, 2)$
$$m_{CD} = \frac{2 – 3}{-3 – 3} = \frac{-1}{-6} = \frac{1}{6}$$
Since $m_{AB} = m_{CD}$, $AB \parallel CD$.
Slope of $BC$, $m_{BC}$: $B(4, 0), C(3, 3)$
$$m_{BC} = \frac{3 – 0}{3 – 4} = \frac{3}{-1} = -3$$
Slope of $AD$, $m_{AD}$: $A(-2, -1), D(-3, 2)$
$$m_{AD} = \frac{2 – (-1)}{-3 – (-2)} = \frac{3}{-1} = -3$$
Since $m_{BC} = m_{AD}$, $BC \parallel AD$.
Since both pairs of opposite sides are parallel, the points are the vertices of a parallelogram.
9. Angle Between $x$-axis and a Line
Find the angle $\theta$ between the $x$-axis and the line joining $A(3, -1)$ and $B(4, -2)$.
The angle the line makes with the positive $x$-axis is related to its slope $m$ by $m = \tan \theta$.
1. Find the Slope $m$
$$m = \frac{y_2 – y_1}{x_2 – x_1} = \frac{-2 – (-1)}{4 – 3} = \frac{-1}{1} = -1$$
2. Find the Angle $\theta$
$$\tan \theta = -1$$
Since $\tan(45^\circ) = 1$, the angle $\theta$ in the second quadrant where $\tan$ is negative is:
$$\theta = 180^\circ – 45^\circ = \mathbf{135^\circ}$$
10. Slopes of Lines with Given Angle
The slope of a line ($m_1$) is double the slope of another line ($m_2$): $m_1 = 2m_2$.
The tangent of the angle $\alpha$ between them is $\frac{1}{3}$: $\tan \alpha = \frac{1}{3}$.
The formula for the tangent of the angle between two lines with slopes $m_1$ and $m_2$ is:
$$\tan \alpha = \left|\frac{m_2 – m_1}{1 + m_1 m_2}\right|$$
Substitute the given values, $m_1 = 2m_2$ and $\tan \alpha = \frac{1}{3}$:
$$\frac{1}{3} = \left|\frac{m_2 – 2m_2}{1 + (2m_2) m_2}\right|$$
$$\frac{1}{3} = \left|\frac{-m_2}{1 + 2m_2^2}\right|$$
This gives two cases, depending on the sign:
Case 1: $\frac{-m_2}{1 + 2m_2^2} = \frac{1}{3}$
$$-3m_2 = 1 + 2m_2^2$$
$$2m_2^2 + 3m_2 + 1 = 0$$
Factor the quadratic: $(2m_2 + 1)(m_2 + 1) = 0$
$$m_2 = -\frac{1}{2} \quad \text{or} \quad m_2 = -1$$
If $m_2 = -\frac{1}{2}$, then $m_1 = 2(-\frac{1}{2}) = -1$.
If $m_2 = -1$, then $m_1 = 2(-1) = -2$.
$$\text{Slopes: } \left(\mathbf{-1, -\frac{1}{2}}\right) \text{ or } \mathbf{(-2, -1)}$$
Case 2: $\frac{-m_2}{1 + 2m_2^2} = -\frac{1}{3}$
$$\frac{m_2}{1 + 2m_2^2} = \frac{1}{3}$$
$$3m_2 = 1 + 2m_2^2$$
$$2m_2^2 – 3m_2 + 1 = 0$$
Factor the quadratic: $(2m_2 – 1)(m_2 – 1) = 0$
$$m_2 = \frac{1}{2} \quad \text{or} \quad m_2 = 1$$
If $m_2 = \frac{1}{2}$, then $m_1 = 2(\frac{1}{2}) = 1$.
If $m_2 = 1$, then $m_1 = 2(1) = 2$.
$$\text{Slopes: } \left(\mathbf{1, \frac{1}{2}}\right) \text{ or } \mathbf{(2, 1)}$$
The possible pairs of slopes for the two lines are $\mathbf{(-1, -1/2)}$, $\mathbf{(-2, -1)}$, $\mathbf{(1, 1/2)}$, and $\mathbf{(2, 1)}$.
11. Slope and Two-Point Form
A line passes through $P(x_1, y_1)$ and $Q(h, k)$. The slope of the line is $m$. Show that $k – y_1 = m (h – x_1)$.
The definition of the slope $m$ of a line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$ is:
$$m = \frac{y_2 – y_1}{x_2 – x_1}$$
Substitute the given points $(x_1, y_1)$ and $(h, k)$ for $(x_2, y_2)$:
$$m = \frac{k – y_1}{h – x_1}$$
Multiply both sides by $(h – x_1)$:
$$m (h – x_1) = k – y_1$$
Rearranging gives the required relation:
$$\mathbf{k – y_1 = m (h – x_1)}$$