Question

The equations to the sides of a triangle are $x-3y=0$ , $4x+3y=0$ and $3x+y=0$ . Then line $3x-4y=0$ passes through
$1)$ The incentre.
$2)$ The centroid.
$3)$ The circumcentre
$4)$ The orthocentre of the triangle

Answer 1

In this type of question we must first find the slopes of each line so that we may know something about the line's behaviour. Then we should find relations between slopes of lines with each other, if there is any relation solution that would be much easier then, otherwise we will have to solve the equations.

Complete step by step answer:
Let first side of the triangle say ${{L}_{1}}$ : $x-3y=0$ ,
So on comparing above line equation with standard line equation that is $y=mx+c$ we get, or we can directly use the formula for slope that is below:
$Slope=\dfrac{-(coefficient\;of\;x)}{(coefficient\;of\;y)}$
Therefore, slope of the line ${{L}_{1}}$ is $\dfrac{1}{3}$
Let call it ${{m}_{1}}=\dfrac{1}{3}$
Let second side of the triangle say ${{L}_{2}}:4x+3y=0$ ,
So on comparing above line equation with standard line equation that is $y=mx+c$ we get, or we can directly use the formula for slope that is below:
$Slope=\dfrac{-(coefficient\;of\;x)}{(coefficient\;of\;y)}$
Therefore, slope of the line ${{L}_{2}}$ is $\dfrac{-4}{3}$
Let call it ${{m}_{2}}=\dfrac{-4}{3}$
Let third side of the triangle say ${{L}_{3}}:3x+y=0$ ,
So on comparing above line equation with standard line equation that is $y=mx+c$ we get, or we can directly use the formula for slope that is below:
$Slope=\dfrac{-(coefficient\;of\;x)}{(coefficient\;of\;y)}$
Therefore, slope of the line ${{L}_{3}}$ is $-3$
Let call it ${{m}_{3}}=-3$ .
Now let the fourth line given is $L:3x-4y=0$ ,
So on comparing above line equation with standard line equation that is $y=mx+c$ we get, or we can directly use the formula for slope that is below:
$Slope=\dfrac{-(coeffient\;of\;x)}{(coefficient\;of\;y)}$
Therefore, slope of the line $L$ is $\dfrac{3}{4}$
Let's call it $m=\dfrac{3}{4}$ .
Now we have completed our first task, now we may try to find any relation between these slopes.
So, if we notice,
${{m}_{1}}\times {{m}_{3}}=-1$ , that is slope of first side of triangle is negative reciprocal of slope of third side of triangle.
So we can say that lines ${{L}_{1}}$ and ${{L}_{3}}$ are perpendiculars.
These are sides of a triangle therefore they form a right angled triangle.
Now if we again notice that slope of second side of triangle that is line ${{L}_{2}}$ is negative reciprocal of slope of fourth line given that is $L$ ,
${{m}_{2}}\times m=-1$
Therefore lines ${{L}_{2}}$ and $L$ are also perpendicular.
Now since line $L$ equation has no intercept it must passes through origin, and since ${{L}_{2}}$ is hypotenuse of the right angled triangle and line $L$ is perpendicular to it, therefore it must passes through the ORTHOCENTRE of the triangle.

So, the correct answer is “Option 4”.

Note: Straight lines are very applicable in real life also not just for virtual problems only. For example, Straight line graphs are used in the research process and the preparation of the government budget and its linear equation can solve many problems like age, money and much more.