Question

find the value of k,if the area of triangle whose vertices are P( k,0),Q(2,2),R(4,3) is 3/2sq. unit.​

Answer 1

Answer:

plz refer to the attached image for your answer

Step-by-step explanation:

hope it helps you

Answer 2

$\huge\underline\bold\orange{Given}$

$\bold{P(k,0)\:Q(2,2)\:R(4,3)}$

$\huge\underline\bold\orange{Formula\: used}$

Area of triangle

$\large\frac{1}{2}{(x1(y2-y3)+x2(y2-y1)+x3(y2-y1)}$

$\huge\underline\bold\orange{Solution}$

$\large\frac{3}{2}{=}\large\frac{1}{2}{(k(2-3)+2(3-0)+4(2-0))}$

$\large\frac{3}{2}{=}$$\large\frac{1}{2}{(k×(-1)+2×3+4×2)}$

$\large\frac{3}{2}{=}$$\large\frac{1}{2}{(-k+6+8)}$

$\large\frac{3×2}{2}{=}$${-k+14}$

${k}{=}{14-3}$

${k}{=}{11}$