Question

If the vertices of the triangie are
(1,k) (4,3) and (-9, 7) and its area
is 15 square unit then find the
value of k​

Answer 1

Answer:

P(k,0)Q(2,2)R(4,3)

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

Formulaused

Area of triangle

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

2

1

(x1(y2−y3)+x2(y2−y1)+x3(y2−y1)

\huge\underline\bold\orange{Solution}

Solution

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

2

3

=

2

1

(k(2−3)+2(3−0)+4(2−0))

\large\frac{3}{2}{=}

2

3

= \large\frac{1}{2}{(k×(-1)+2×3+4×2)}

2

1

(k×(−1)+2×3+4×2)

\large\frac{3}{2}{=}

2

3

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

2

1

(−k+6+8)

\large\frac{3×2}{2}{=}

2

3×2

= {-k+14}−k+14

{k}{=}{14-3}k=14−3

{k}{=}{11}k=11

Answer 2

Answer:

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