Question

The vertices of ∆ ABC are A(-2, 1), B(5, 4) and C(2, -3) respectively. Find the area of triangle and

height along the side AC.

pls solve neatly and clearly on white sheet​

Answer 1

$Given \: the \: vertices \:of \: \triangle \: ABC \\are\: A(-2 ,1 ) , \: B(5,4) \: and \: C(2, -3)$

$Let \: A(-2 , 1 ) = ( x_{1} , y_{1} ) , \\B(5 , 4) = ( x_{2} , y_{2} ) ,\\and \: C(2 , -3) = ( x_{3} , y_{3} )$

$1.\red{Area \: of \: \triangle ABC } \\= \frac{1}{2}|x_{1}(y_{2}-y_{3}) + x_{2}(y_{3}-y_{1}) + x_{3}(y_{2}-y_{1}) |$

$= \frac{1}{2}|-2[4-(-3)]+5(-3-1)+2(1-4)|\\= \frac{1}{2}|(-2)\times 7 + 5\times (-4) + 2 \times (-3)| \\= \frac{1}{2} | -14 - 20 - 6 | \\= \frac{1}{2} | -40| \\= \frac{40}{2} \\ \green {= 20 \: square \:units} \: --(1)$

$2.\blue { Base \:of \: triangle \: ABC } \\ \red{Distance\: between \: A \: and \: C} \\= \sqrt{ ( x_{3} - x_{1} )^{2} + ( y_{3} - y_{1} )^{2} } \\= \sqrt{ [ 2- (-2) ]^{2} + ( -3 - 1 )^{2} } \\= \sqrt{ (2+2)^{2} + (-4)^{2} } \\= \sqrt{ 16 + 16 } \\= \sqrt{ 32 } \\= 4 \sqrt{2} \: --(2)$

$Let \: \orange { Corresponding \: height = h }$

$3. Now, Area \: of \: \triangle ABC = 20$

$\implies \pink { \frac{1}{2} \times AC \times h } = 20$

$\implies \frac{1}{2} \times 4\sqrt{2} \times h = 20$

$\implies 2\sqrt{2} \times h = 20$

$\implies h = \frac{ 20 }{2\sqrt{2}}$

$\implies h = \frac{ 10 }{\sqrt{2}}$

$\implies h = \frac{ 10 \sqrt{2}}{\sqrt{2}\times \sqrt{2}}$

$\implies h = \frac{ 10 \sqrt{2}}{2}$

$\implies h = 5\sqrt{2}$

Therefore.,

$1.\red{Area \: of \: \triangle ABC }\\\green {= 20 \: square \:units}$

$2. \red{ Height \:along \: side \: AC } \green { = 5\sqrt{2} \: units }$

♪••.