Question

Question 17
If the points A(2,9), B(a,5) and C(5,5) are the vertices of a $\Delta$ ABC right angled at B, then find the values of a and hence the area of $\Delta$ ABC.

Answer 1

Given that the line segment joining the points A(2,9), B(a,5) and C(5,5) are the vertices of a $\Delta$ ABC right angled at B.
By Pythagoras theorem, $A{C}^2=A{B}^2+B{C}^2$ ...(i)
Now, by distance formula,
$AB=\sqrt{(a-2{)}^2+(5-9{)}^2}\left[\begin{array}{c}\\ \because distancebetweentwopoints(x_1,y_1)and(x_2,y_2),d\\ =\sqrt{(x_2-x_1)+(y_2-y_1{)}^2}\end{array}\right]=\sqrt{a^2+4-4a+16}=\sqrt{a^2-4a+20}BC=\sqrt{(5-a{)}^2+(5-5{)}^2}=\sqrt{(5-a{)}^2+0}=5-aAndAC=\sqrt{(2-5{)}^2+(9-5{)}^2}=\sqrt{(-3{)}^2+(4)}=\sqrt{9+16}=\sqrt{25}=5PutthevaluesofAB,BCandACinEq.\left(i\right),weget(5{)}^2=(\sqrt{a^2-4a+20}{)}^2+(5-a{)}^2\Rightarrow 25=a^2-4a+20+25+a^2-10a\Rightarrow 2a^2-14a+20=0\Rightarrow a^2-7a+10=0\Rightarrow a^2-2a-5a+10=0[byfactorizationmethod]\Rightarrow a(a-2)-5(a-5)=0\Rightarrow (a-2\left)\right(a-5)=0\therefore a=2,5$
$\text{Here, a}\ne \text{5, because if a=5, the length of BC becomes zero (0)}$
$\text{which is not possible because the sides AB, BC and CA form a right}$
$\text{angled triangle.}$

Now, the coordinates of A, B and C become (2,9), (2,5) and (5,5) respectively.
$\because Areaof\Delta ABC=\Delta =\frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]\therefore \Delta =\frac{1}{2}\left[2\right(5-5)+2(5-9)+5(9-5\left)\right]=\frac{1}{2}[2\times 0+2(-4)+5(4\left)\right]=\frac{1}{2}(0-8+20)=\frac{1}{2}\times 12=6$
Hence, the required area of $\Delta$ABC is 6 sq units.