Question

The area of a triangle is 20 sq. units. Two of its vertices are (4, 2) and (6, −4). If the third vertex is (7, a), find the values of a.

Answer 1

$\underline{\dag\sf \ \ Given :- \ \ \ }$

$\bullet\sf\ \ Area \ of \ triangle= 20sq.units$

$\sf{Vertices\ of\ \Delta }\begin{cases}\sf{\ (4,2)}\\ \sf{\ (6, -4)}\\ \sf{\ (7,a)}\end{cases}$

$\underline{\dag\sf \ \ To \ Find :- \ \ \ }$

• We have to find the value of a in the third vertex of triangle .

$\underline{\mathbb{\ \ \ SOLUTION}}$

$\underline{ \large{\boxed{ \sf{Area \ of \ \Delta= \dfrac{1}{2}{\huge\mid}x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2){\huge\mid}}}}}$

$\sf{We\ Have}\begin{cases}\sf{\ x_1= 4\ \ ;\ x_2=6\ \ ;\ x_3=7}\\ \sf{\ y_1=2\ \ ;\ y_2= (-4)\ \ ;\ y_3=a}\end{cases}$

Now put these values in the formula -

$:\implies\sf 20= \dfrac{1}{2}{\huge\mid}4(-4-a)+6(a-2)+7[2-(-4)]{\huge\mid}\\ \\ \\ \\ :\implies\sf 20\times 2= {\huge\mid}-16-4a+6a-12+42{\huge\mid}\\ \\ \\ \\ :\implies\sf 40= {\huge\mid} 42-28+2a{\huge\mid}\\ \\ \\ \\ :\implies\sf 40 = {\huge\mid}14+2a{\huge\mid}\\ \\ \\ \\ :\implies\sf 40-14=2a\\ \\ \\ \\ :\implies\sf 26=2a\\ \\ \\ \\ :\implies\sf \cancel{\dfrac{26}{2}}=a\\ \\ \\ \\ :\implies\sf a= \underline{\boxed{\purple{\sf 13 \: unit}}}$

$\underline{\large {\boxed{\sf \: Third \ vertex\ of \ \Delta=\mathfrak{(7,13)}}}}$

Answer 2

Area=20

Vertices= (4,2) , (6,-4),(7,a)

Now according to the formula -

Area = 1/2(x1(y2-y3)+x2(y3-y1)+x3(y1-y2)

→ 20 =1/2( 4(4-a)+6(a-2)+7(2+4)

→40= 16-4a+6a-12+42

40-14=2a

a=13

the value of a is 13