Question

The coordinates of the orthocenter of the triangle whose sides are represented by the lines \(4x-7y+10 = 0\), \(x+y = 5\) and \(7x+4y = 15\) are \(\ldots\)

• A. \((1,2)\)
• B. \((1,-2)\)
• C. \((-1,2)\)
• D. \((-1,-2)\)

Hint:

Check whether two sides are perpendicular. If so, the vertex between them is the orthocentre.

Answer 1

The Correct Option is A

Step 1: Find the vertices and test options:
Let the lines be $L_1: 4x-7y+10=0$, $L_2: x+y=5$, $L_3: 7x+4y=15$.

Step 2: Use the right angle:
The slopes of $L_1$ and $L_3$ are $\frac47$ and $-\frac74$, whose product is $-1$. So the angle at $L_1\cap L_3$ is $90^\circ$.

Step 3: Solve for that vertex:
From $L_3$: $y = \frac{15-7x}{4}$. Put in $L_1$: $4x - \frac{7(15-7x)}{4} + 10 = 0$, so $16x - 105 + 49x + 40 = 0$, so $65x = 65$ and $x=1$, $y = 2$.
The right-angle vertex is the orthocentre, $(1,2)$. Check: $(1,2)$ satisfies $4 - 14 + 10 = 0$ and $7 + 8 = 15$.

Final Answer:
$(1,2)$, option (A). \[ \boxed{(1,2) \text{ (A)}} \]