Circles Question - None

Question

Let $A B C$ be the triangle such that the equations of lines $A B$ and $A C$ be $3 y-x=2$ and $x+y=2$, respectively, and the points $B$ and $C$ lie on $x$-axis. If $P$ is the orthocentre of the triangle $A B C$, then the area of the triangle $P B C$ is equal to

Accepted Answer

$x=1$
We are given triangle $A B C$ with:

Line AB : $3 y-x=2$
Line AC: $x+y=2$
Points B and C lie on the x-axis.
$P$ is the orthocenter of triangle $A B C$.

The point A is the intersection of lines AB and AC, A(1, 1)
The orthocentre is the intersection point of the altitudes of the triangle.
Altitude from A to BC:

Since BC lies on the $x$-axis, the altitude from A is a vertical line passing through A.
Equation: $x=1$

Altitude from B to AC:

Slope of AC: From $x+y=2$, slope $m_{{AC}}=-1$
Slope of altitude BP: Perpendicular to ${AC}, {so}\ m_{{BP}}=1$.
Equation of BP: $y+0=1(x+2)$ or $y=x+2$.

Find ${P}$:

Intersection of $x=1$ and $y=x+2$ :
$y=1+2=3$
Thus, ${P}(1,3)$.

Using the area formula for a triangle:

Area $=\frac{1}{2} imes$ base × height

$=\frac{1}{2} imes 4 imes 3=6$

Question:

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