Question

ABC is on isosceles triangle of
25/6 square units. If the
coordinates of the base are
B(1,3) ond C(-2,7), then the
coordinates of A are
Clear Response​

Answer 1

Answer:

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Answer 2

Step-by-step explanation:

Find the length of the base:

Given that, the coordinates of the base are $B (1, 3)$ and $C (- 2, 7)$.

Thus the length of $BC$

$=\sqrt{(1+2)^{2}+(3-7)^{2}}$ units

$=\sqrt{3^{2}+4^{2}}$ units

$=\sqrt{9+16}$ units

$=\sqrt{25}$ units

$5$ units

Find the length of the height:

Since the area of the given isosceles triangle $ABC$ is $\frac{25}{6}$ square units, the length of the height $AD$ (drawn) be

$\quad 2\times\frac{area\:of\:\Delta ABC}{length\:of\:the\:base}$

$=2\times\frac{\frac{25}{6}}{5}$ units

$=2\times\frac{25}{6\times 5}$ units

$=\frac{50}{30}$ units

$=\frac{5}{3}$ units

Find the equation of the base:

The equation of the base $BC$ is

$\quad \frac{y-7}{7-3}=\frac{x+2}{-2-1}$

$\Rightarrow \frac{y-7}{4}=\frac{x+2}{-3}$

$\Rightarrow 4x+8=-3y+21$

$\Rightarrow 4x+3y=13$ ___(1)

Find the mid-point of the base:

The coordinates of the mid-point $D$ of the base $BC$ are

$\quad (\frac{1-2}{2},\frac{3+7}{2})$ i.e., $(-\frac{1}{2},5)$

Find the equation of the height:

Since $AD$ is perpendicular to $BC$, let the equation of $AD$ be

$\quad 3x-4y=k$ where $k$ is constant

Since $AD$ pαsses through the point $D(-\frac{1}{2},5)$,

$\quad 3(-\frac{1}{2})-4(5)=k$

$\Rightarrow k=-\frac{43}{2}$

Thus the equation of $AD$ is

$\quad 3x-4y=-\frac{43}{2}$

$\Rightarrow 6x-8y=-43$ ___ (2)

Consider the vertex $A$:

Let the coordinates of $A$ be $(m,n)$.

Satisfy equation no. (2) with $A$:

Since $AD$ is satisfied by $A(m,n)$,

$\quad 6m-8n=-43$ ___ (3)

Use the distance of $A$ from the base:

Here the distance of $A(m,n)$ from the base $BC:4x+3y=13$ is $\frac{5}{3}$ units. Then

$\quad \frac{4m+3n-13}{\sqrt{4^{2}+3^{2}}}=\frac{5}{3}$

$\Rightarrow \frac{4m+3n-13}{5}=\frac{5}{3}$

$\Rightarrow 12m+9n-39=25$

$\Rightarrow 12m+9n=64$ ___ (4)

Find the value of $m$ and $n$:

We have two equations to solve,

$\quad 6m-8n=-43\quad\quad\times 2$

$\quad 12m+9n=64\quad\quad\times 1$

$\Rightarrow$

$\quad 12m-16n=-86$

$\quad 12m+9n=64$

On subtraction, we get

$\quad 25n=150\Rightarrow n=6$

Thus $m=\frac{5}{6}$

Answer:

$\therefore$ the coordinates of $A$ are $(\frac{5}{6},6)$.