# 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.

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 $(3, 4).$