Question

35. For what value of k, (k 0), is the area of the triangle with vertices (-2, 5), (k, -4) and (2k + 1, 10) equal to 53 sq. units​

Answer 1

ANSWER:

Given:

• Vertices of triangle = (-2, 5), (k, -4) and (2k + 1, 10).
• Area of triangle = 53 square units

To Find:

• Value of k

Diagram:

$\setlength{\unitlength}{1 cm}\begin{picture}(0,0)\thicklines\qbezier(1, 0)(1,0)(3,3)\qbezier(5,0)(5,0)(3,3)\qbezier(5,0)(1,0)(1,0)\put(2.5,3.2){ \bf A(-2, 5) }\put(0.2,-0.34){ \bf C(2k + 1, 10) }\put(5,-0.2){ \bf B(k, -4) }\end{picture}$

Solution:

$\text{We are given that,}\\\\:\longrightarrow\text{Area of \Delta ABC = 53 square units}.\\\\\text{We know that,}\\\\:\hookrightarrow\text{Area of a triangle}=\dfrac{1}{2}\bigg|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\bigg|\\\\\text{So,}\\\\\text{In this question,}\\\\:\longrightarrow x_1=-2, y_1=5, x_2=k, y_2=-4, x_3=2k+1, y_3=10.\\\\\text{So,}$

$:\implies\text{Area of \Delta ABC}=\dfrac{1}{2}\bigg|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\bigg|\\\\:\implies\text{Area of \Delta ABC}=\dfrac{1}{2}\bigg|(-2)(-4-10)+(k)(10-5)+(2k+1)(5-(-4))\bigg|\\\\:\implies\text{Area of \Delta ABC}=\dfrac{1}{2}\bigg|(-2)(-14)+(k)(5)+(2k+1)(5+4)\bigg|$

$:\implies\text{Area of \Delta ABC}=\dfrac{1}{2}\bigg|28+5k+18k+9\bigg|\\\\:\implies\text{Area of \Delta ABC}=\dfrac{1}{2}\bigg|23k+37\bigg|\\\\\text{As, Area is 53,}\\\\:\implies53=\dfrac{1}{2}\bigg|23k+37\bigg|\\\\:\implies106=\bigg|23k+37\bigg|\\\\\text{As, k > 0,}\\\\:\implies106=23k+37\\\\:\implies23k=106-37\\\\:\implies23k=69\\\\:\implies k=69/23\\\\\bf{:\implies k=3}\\\\\text{\bf{Hence, value of k is 3units.}}$

Formula Used:

$:\hookrightarrow\text{Area of a triangle}=\dfrac{1}{2}\bigg|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\bigg|$