Two sides of a rhombus are along the lines, x - y + 1 = 0 and 7x - y - 5 = 0 . If its diagonals intersect at (-1, -2) , then which one of the following is a vertex of this rhombus?

Straight Lines and Pair of Straight Lines PYQ — JEE Main 2016 · Shift 0 · Q33 with Solution

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Correct answer: A

Interpret the given information

Two sides of the rhombus lie on the lines $L_1: x-y+1=0$ $L_2: 7x-y-5=0$ These two sides must meet at a vertex of the rhombus.

The diagonals intersect at the center of the rhombus: $O=(-1,-2)$ In a rhombus, diagonals bisect each other, so opposite vertices are symmetric about the center.

Find the common vertex of the given side lines

Solve $x-y+1=0 \Rightarrow y=x+1$ $7x-y-5=0 \Rightarrow y=7x-5$ Equating: $x+1=7x-5$ $6=6x$

\quad y=2$$ So one vertex is $$A=(1,2)$$ --- 3. **Find the opposite vertex using the center** If $O$ is the midpoint of diagonal joining $A$ and the opposite vertex $C$, then $$O=\left(\frac{x_A+x_C}{2},\frac{y_A+y_C}{2}\right)$$ So $$x_C=2(-1)-1=-3$$ $$y_C=2(-2)-2=-6$$ Hence the opposite vertex is $$C=(-3,-6)$$ --- 4. **Let the other two vertices be $B$ and $D$** In a parallelogram (hence in a rhombus), diagonals bisect each other, so $$B+D=2O=(-2,-4)$$ Also, for a rhombus, all sides are equal. Since $A$ is connected to $B$ along one given line and to $D$ along the other given line, the points $B$ and $D$ must lie on those lines respectively, and satisfy $$AB=AD$$ A simpler vector method: If adjacent side vectors from $A$ are $\vec u$ and $\vec v$, then opposite vertex is $$C=A+\vec u+\vec v$$ Thus $$\vec u+\vec v=C-A=(-3,-6)-(1,2)=(-4,-8)$$ Now, - $\vec u$ is along $L_1: x-y+1=0$, whose direction vector is $(1,1)$. - $\vec v$ is along $L_2: 7x-y-5=0$, whose direction vector is $(1,7)$. So let $$\vec u=\lambda(1,1), \quad \vec v=\mu(1,7)$$ Then $$\lambda(1,1)+\mu(1,7)=(-4,-8)$$ This gives $$\lambda+\mu=-4$$ $$\lambda+7\mu=-8$$ Subtracting, $$6\mu=-4 \Rightarrow \mu=-\frac23$$ Then $$\lambda=-4+\frac23=-\frac{10}{3}$$ Thus $$\vec u=\left(-\frac{10}{3},-\frac{10}{3}\right), \quad \vec v=\left(-\frac23,-\frac{14}{3}\right)$$ Therefore the adjacent vertices are $$B=A+\vec u=(1,2)+\left(-\frac{10}{3},-\frac{10}{3}\right)=\left(-\frac73,-\frac43\right)$$ $$D=A+\vec v=(1,2)+\left(-\frac23,-\frac{14}{3}\right)=\left(\frac13,-\frac83\right)$$ So one vertex of the rhombus is $$\left(\frac13,-\frac83\right)$$ which matches option A. --- 5. **Check options** - **A**: $\left(\frac13,-\frac83\right)$ ✓ obtained above. - **B**: notation is malformed, but it does not match any actual vertex. - **C**: $(-3,-9)$ is not a vertex. - **D**: $(-3,-8)$ is not a vertex. Hence the correct option is **A**. --- 6. **Comparison with stored answer** Stored correct answer: **A** Our derived answer: **A** They agree.

More from Straight Lines and Pair of Straight Lines

View written solutionFree

Correct answer: A

Interpret the given information

Two sides of the rhombus lie on the lines $L_1: x-y+1=0$ $L_2: 7x-y-5=0$ These two sides must meet at a vertex of the rhombus.

The diagonals intersect at the center of the rhombus: $O=(-1,-2)$ In a rhombus, diagonals bisect each other, so opposite vertices are symmetric about the center.

Find the common vertex of the given side lines

Solve $x-y+1=0 \Rightarrow y=x+1$ $7x-y-5=0 \Rightarrow y=7x-5$ Equating: $x+1=7x-5$ $6=6x$

\quad y=2$$ So one vertex is $$A=(1,2)$$ --- 3. **Find the opposite vertex using the center** If $O$ is the midpoint of diagonal joining $A$ and the opposite vertex $C$, then $$O=\left(\frac{x_A+x_C}{2},\frac{y_A+y_C}{2}\right)$$ So $$x_C=2(-1)-1=-3$$ $$y_C=2(-2)-2=-6$$ Hence the opposite vertex is $$C=(-3,-6)$$ --- 4. **Let the other two vertices be $B$ and $D$** In a parallelogram (hence in a rhombus), diagonals bisect each other, so $$B+D=2O=(-2,-4)$$ Also, for a rhombus, all sides are equal. Since $A$ is connected to $B$ along one given line and to $D$ along the other given line, the points $B$ and $D$ must lie on those lines respectively, and satisfy $$AB=AD$$ A simpler vector method: If adjacent side vectors from $A$ are $\vec u$ and $\vec v$, then opposite vertex is $$C=A+\vec u+\vec v$$ Thus $$\vec u+\vec v=C-A=(-3,-6)-(1,2)=(-4,-8)$$ Now, - $\vec u$ is along $L_1: x-y+1=0$, whose direction vector is $(1,1)$. - $\vec v$ is along $L_2: 7x-y-5=0$, whose direction vector is $(1,7)$. So let $$\vec u=\lambda(1,1), \quad \vec v=\mu(1,7)$$ Then $$\lambda(1,1)+\mu(1,7)=(-4,-8)$$ This gives $$\lambda+\mu=-4$$ $$\lambda+7\mu=-8$$ Subtracting, $$6\mu=-4 \Rightarrow \mu=-\frac23$$ Then $$\lambda=-4+\frac23=-\frac{10}{3}$$ Thus $$\vec u=\left(-\frac{10}{3},-\frac{10}{3}\right), \quad \vec v=\left(-\frac23,-\frac{14}{3}\right)$$ Therefore the adjacent vertices are $$B=A+\vec u=(1,2)+\left(-\frac{10}{3},-\frac{10}{3}\right)=\left(-\frac73,-\frac43\right)$$ $$D=A+\vec v=(1,2)+\left(-\frac23,-\frac{14}{3}\right)=\left(\frac13,-\frac83\right)$$ So one vertex of the rhombus is $$\left(\frac13,-\frac83\right)$$ which matches option A. --- 5. **Check options** - **A**: $\left(\frac13,-\frac83\right)$ ✓ obtained above. - **B**: notation is malformed, but it does not match any actual vertex. - **C**: $(-3,-9)$ is not a vertex. - **D**: $(-3,-8)$ is not a vertex. Hence the correct option is **A**. --- 6. **Comparison with stored answer** Stored correct answer: **A** Our derived answer: **A** They agree.

Straight Lines and Pair of Straight Lines question

Straight Lines and Pair of Straight Lines question

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Straight Lines and Pair of Straight Lines question

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