Straight Lines

Class 11 NCERT

NCERT

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 :

Let $ABCD$ be the given quadrilateral with vertices $A (-4, 5), B (0, 7), C (5, -5),$ and $D (-4, - 2).$$\\$ Then, by plotting $A, B, C$ and $D$ on the Cartesian plane and joining $AB, BC, CD,$ and $DA,$ the given quadrilateral can be drawn as$\\$

To find the area of quadrilateral$ ABCD,$ we draw one diagonal, say $AC.$ Accordingly, area $(ABCD) =$ area $( \Delta ABC) $+ area $( \Delta ACD)$ We know that the area of a triangle whose vertices are $(x _1 , y_ 1 ), (x_ 2 , y_ 2 ),$ and $(x_ 3 , y _3 )$ is$\\$ $\dfrac{1}{2}|x_1(y_2-y_3)+(y_3-y_1)+x_3(y_1-y_2)|$$\\$ Therefore, area of $\Delta ABC$$\\$ $=\dfrac{1}{2}|-4(7+5)+0(-5-5)+5(5-7) \text{unit^2}|\\ =\dfrac{1}{2}|-4(12)+5(-2)|unit^2\\ =\dfrac{1}{2}|-48-10|unit^2\\ \dfrac{1}{2}|-58|unit^2\\ =\dfrac{1}{2}*58 unit^2\\ 29 unit^2$$\\$ Area of $ \Delta ACD$$\\$ $=\dfrac{1}{2}|-4(-5+2)+5(-2-5)+(-4)(5-5)|unit^2\\ =\dfrac{1}{2}|-4(-3)+5(-7)-4(10)|unit^2\\ =\dfrac{1}{2}|12-35-40|unit^2\\ =\dfrac{1}{2}|-63| unit^2\\ \dfrac{63}{2}unit^2$ Thus, area $(ABCD) =(29+\dfrac{63}{2})unit^2\\ =\dfrac{58+63}{2}unit^2=\dfrac{121}{2}unit^2$

2   The base of an equilateral triangle with side $2a$ lies along the y-axis such that the midpoint of the base is at the origin. Find vertices of the triangle.

Solution :

Let $ABC$ be the given equilateral triangle with side $2a.$$\\$ Accordingly,$AB = BC = CA = 2a$ Assume that base BC lies along the y-axis such that the mid-point of $BC$ is at the origin. i.e., $BO = OC = a,$ where $O$ is the origin. Now, it is clear that the coordinates of point C are $(0, a),$ while the coordinates of point B are $(0, -a).$$\\$ It is known that the line joining a vertex of an equilateral triangle with the mid-point of its opposite side is perpendicular. Hence, vertex A lies on the y-axis.$\\$

On applying Pythagoras theorem to $\Delta AOC,$ we obtain$\\$ $(AC)^ 2 =( OA)^2+( OC )^2\\ \implies (2a)^2=(OA)^2+a^2\\ \implies 4a^2-a^2=(OA)^2\\ \implies (OA)^2=3a^2\\ \implies OA=3a$ $\therefore $ Coordinates of point $A = (\pm 3 a , 0)$ Thus, the vertices of the given equilateral triangle are $(0,a ) , ( 0, - a ) ,$ and $(\sqrt{3a},0) \ or \ (0,a),(0,-a),$ and $(-\sqrt{3a},0)$

3   Find the distance between $P ( x _1 , y_ 1 )$ and $Q ( x _2 , y _2 )$ when $\\$(i) PQ is parallel to the y-axis, $\\$(ii) PQ is parallel to the x-axis.

Solution :

The given points are $P ( x_ 1 , y_ 1 )$ and $Q (x _2 , y _2 ) .$$\\$ (i) When PQ is parallel to the y-axis,$ x _1 = x_ 2 .$$\\$ In this case, distance between P and Q =$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\ =\sqrt{(y_2-y_1)^2}\\ =|y_2-y_1|$$\\$ (ii)In this case, distance between P and Q =$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\ =\sqrt{(x_2-x_1)^2}\\ =|x_2-x_1|$

4   Find a point on the x-axis, which is equidistant from the points $(7, 6)$ and $(3, 4).$

Solution :

Let $(a,0)$ be the point on the X- axis that is equidistance from the points $(7, 6)$ and $(3, 4).$$\\$ Accordingly, $\sqrt{(7-a)^2+(6-0)^2}\\ =\sqrt{(3-a)^2+(4-0)^2}\\ \implies \sqrt{49+a^2-14a+36}\\ =\sqrt{9+a^2-6a+16}\\ \implies \sqrt{a^2-14a+85}\\ =\sqrt{a^2-6a+25}$$\\$ On squaring both sides, we obtain$\\$ $a^2-14a+85=a^2-6a+25\\ \implies -14a+6a=25-85\\ \implies -8a=-60\\ \implies a=\dfrac{60}{8}=\dfrac{15}{2}$$\\$ Thus, the required point on the x-axis is $(\dfrac{15}{2},0)$

5   Find the slope of a line, which passes through the origin, and the mid-point of the segment joining the points $P (0, -4)$ and $B (8, 0).$

Solution :

The coordinates of the mid-point of the line segment joining the points $P (0, -4)$ and $B (8, 0)$ are $(\dfrac{0+8}{2},\dfrac{-4+0}{2})=(4,-2)$$\\$ It is known that the slope (m) of a non-vertical line passing through the points $(x_ 1 , y_ 1 )$ and $(x_ 2 ,y_2)$ is given by $ m=\dfrac{y_2-y_1}{x_2-x_1},x_2\neq x_1$$\\$ Therefore, the slope of the line passing through $(0, 0,)$ and $(4, -2)$ is $\dfrac{-2-0}{4-0}=\dfrac{-2}{4}=\dfrac{-1}{2}$ Hence, the required slope of the line is $-\dfrac{1}{2}$

6   Without using the Pythagoras theorem, show that the points $(4, 4), (3, 5)$ and $(-1, -1)$ are vertices of a right angled triangle.

Solution :

The vertices of the given triangle are $A (4, 4), B (3, 5),$ and $C (-1, -1).$$\\$ It is known that the slope $(m)$ of a non-vertical line passing through the points $(x_ 1 , y_ 1 )$ and $(x_2,y_2)$ is given by $ m=\dfrac{y_2-y_1}{x_2-x_1},x_2 \neq x_1$$\\$ $\therefore $ Slope of AB $(m_1)=\dfrac{5-4}{3-4}=-1$$\\$ Slope of BC$(m_2)=\dfrac{-1-5}{-1-3}=\dfrac{-6}{-4}=\dfrac{3}{2}$$\\$ Slope of CA$(m_3)=\dfrac{4+1}{4+1}=\dfrac{5}{5}=1$$\\$ It is observed that $m _1 m _3 = -1$$\\$ This shows that line segments AB and CA are perpendicular to each other i.e., the given triangle is right-angled at $A (4, 4).$$\\$ Thus, the points $( 4, 4 ) , (3,5 )$, and $(-1, -1)$ are the vertices of a right-angled triangle.

7   Find the slope of the line, which makes an angle of $30 ^o$ with the positive direction of y-axis measured anticlockwise.

Solution :

If a line makes an angle of $30^o$ with positive direction of the y-axis measured anticlockwise, then the angle made by the line with the positive direction of the x-axis measured anticlockwise is $90^o + 30^o 120 ^o$ . Thus, the slope of the given line is $\tan 120 ^o = \tan (180 ^o60^o)=-\tan 60^o=-\sqrt{ 3}$

8   Find the value of $x$ for which the points $(x, -1), (2, 1)$ and $(4, 5)$ are collinear.

Solution :

If points $A (x, -1), B (2, 1),$ and $C (4, 5)$ are collinear, then Slope of AB = Slope of BC$\\$ $\Rightarrow \dfrac{1-(-1)}{2-x}=\dfrac{5-1}{4-2}\\ \Rightarrow \dfrac{1+1}{2-x}=\dfrac{4}{2}\\ \Rightarrow \dfrac{2}{2-x}=2\\ \Rightarrow 2=4-2x\\ \Rightarrow 2x=2\\ \Rightarrow x=1$$\\$ Thus, the required value of $x$ is $1.$

9   Without using distance formula, show that points $(-2, - 1 ) , ( 4,0 ) , ( 3,3 )$ and $(- 3,2 )$ are vertices of a parallelogram.

Solution :

Let points $(-2, -1 ), ( 4,0 ) , ( 3,3 )$ and $(-3,2 )$ be respectively denoted by $A, B, C,$ and $D.$$\\$

Slopes of AB=$\dfrac{0+1}{4+2}=\dfrac{1}{6}$$\\$ Slope of CD$=\dfrac{2-3}{-3-3}=\dfrac{-1}{-6}=\dfrac{1}{6}$$\\$ $\Rightarrow $ Slope of AB =Slope of CD$\\$ $\Rightarrow $ AB and CD are parallel to each other.$\\$ Now , slope of BC$=\dfrac{3-0}{3-4}=\dfrac{3}{-1}=-3$$\\$ Slope of AD$= \dfrac{2+1}{-3+2}=\dfrac{3}{-1}=-3$$\\$ $\Rightarrow $ Slope of BC = Slope of AD$\\$ $\Rightarrow $ BC and AD are parallel to each other $\\$ Therefore, both pairs of opposite side of quadrilateral $ABCD$ are parallel. Hence, $ABCD$ is a parallelogram.$\\$ Thus, points $(- 2, - 1 ), ( 4, 0 ) , ( 3,3 )$ and $( -3, 2 )$ are the vertices of a parallelogram.

10   Find the angle between the x-axis and the line joining the points $(3, -1)$ and $(4, -2)$.

Solution :

The slope of the line joining the points $(3,-1)$ and $ (4,-2)$ is $ m=\dfrac{-2-(-1)}{4-3}=-2+1=-1$$\\$ Now, the inclination ( $\theta $ ) of the line joining the points $(3, -1)$ and $(4, -2)$ is given by $\tan \theta = -1$$\\$ $\Rightarrow \theta =(90^o+45^o)=135^o$$\\$ Thus, the angle between the x-axis and the line joining the points $(3, -1) $ and $(4, -2)$ is $135 ^o$