Two equal sides of an isosceles triangle are along -x + 2y = 4 and x + y = 4 . If m is the slope of its third side, then the sum, of all possible distinct values of m , is:

Straight Lines and Pair of Straight Lines PYQ — JEE Main 2025 · 28 Jan · Shift 2 · Q31 with Solution

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Correct answer: D

Given equal sides

The two equal sides of the isosceles triangle lie along $L_1:",-x+2y=4 \quad \Rightarrow \quad y=\frac{x}{2}+2$ and $L_2:",x+y=4 \quad \Rightarrow \quad y=-x+4.$

These two lines meet at the vertex where the equal sides of the isosceles triangle intersect.

Find the vertex of intersection

Solve $-x+2y=4, \qquad x+y=4.$

From $x+y=4$ , we get $x=4-y$ . Substitute into the first equation: $-(4-y)+2y=4$ $-4+y+2y=4$ $3y=8 \Rightarrow y=\frac{8}{3}.$ Then $x=4-\frac{8}{3}=\frac{4}{3}.$

So the common vertex is $A\left(\frac{4}{3},\frac{8}{3}\right).$

Angle bisectors of the two equal sides

If the two equal sides are along $L_1$ and $L_2$ , then the third side (base) must be perpendicular to one of the angle bisectors at the vertex, because in an isosceles triangle the altitude from the apex is also the angle bisector and is perpendicular to the base.

So first find the angle bisectors of the pair of lines.

Write lines in standard form: $L_1: -x+2y-4=0, \qquad L_2: x+y-4=0.$

Angle bisectors satisfy $\frac{-x+2y-4}{\sqrt{(-1)^2+2^2}}=\pm \frac{x+y-4}{\sqrt{1^2+1^2}}$ $\frac{-x+2y-4}{\sqrt5}=\pm \frac{x+y-4}{\sqrt2}.$

Find slopes of the angle bisectors

Instead of solving fully, use the fact that if a bisector has slope $s$ , then it makes equal angles with the two given lines.

The slopes of the given lines are $m_1=\frac12, \qquad m_2=-1.$

The angle between a bisector and each line is equal, so the two bisectors are the internal and external bisectors of these directions. It is simpler to get their slopes directly from the combined equations.

Case 1:

$\sqrt2(-x+2y-4)=\sqrt5(x+y-4).$ Rearrange: $(-\sqrt2-\sqrt5)x+(2\sqrt2-\sqrt5)y+(-4\sqrt2+4\sqrt5)=0.$ Thus slope is $s_1=-\frac{-\sqrt2-\sqrt5}{2\sqrt2-\sqrt5}=\frac{\sqrt2+\sqrt5}{2\sqrt2-\sqrt5}.$ Rationalizing / simplifying gives $s_1=4+3\sqrt{10}.$

Case 2:

$\sqrt2(-x+2y-4)=-\sqrt5(x+y-4).$ Rearrange: $(-\sqrt2+\sqrt5)x+(2\sqrt2+\sqrt5)y+(-4\sqrt2-4\sqrt5)=0.$ Thus slope is $s_2=-\frac{-\sqrt2+\sqrt5}{2\sqrt2+\sqrt5}=\frac{\sqrt2-\sqrt5}{2\sqrt2+\sqrt5}.$ Simplifying, $s_2=4-3\sqrt{10}.$

So the two angle bisectors have slopes $4+3\sqrt{10}, \qquad 4-3\sqrt{10}.$

Slope of the third side

The third side is perpendicular to an angle bisector, so if its slope is $m$ , then $m=-\frac1s.$

Hence possible slopes are

\qquad m_2=-\frac{1}{4-3\sqrt{10}}.$$ Now add them: $$m_1+m_2=-\left(\frac{1}{4+3\sqrt{10}}+\frac{1}{4-3\sqrt{10}}\right).$$ Using $$\frac{1}{a+b}+\frac{1}{a-b}=\frac{2a}{a^2-b^2},$$ with $a=4$, $b=3\sqrt{10}$, $$m_1+m_2=-\frac{2\cdot 4}{16-90} =-\frac{8}{-74}=\frac{4}{37}.$$ This looks inconsistent with the options, so let us use a cleaner geometric approach. --- 6. **Cleaner approach using angle formula** Let the apex angle between the equal sides be $\theta$. Then the base makes equal angles with the two sides. If the base has slope $m$, then the perpendicular from apex to base is an angle bisector. Let $\phi_1=\tan^{-1}(1/2)$ and $\phi_2=\tan^{-1}(-1)$ be inclinations of the equal sides. The bisector inclinations are $$\frac{\phi_1+\phi_2}{2}, \qquad \frac{\phi_1+\phi_2}{2}+90^\circ.$$ The base inclinations are each $90^\circ$ from these, i.e. again the pair of lines whose slopes satisfy the quadratic obtained from angle-bisector relation. A more direct algebraic route is to note: For a line of slope $m$ to cut the two given lines at points equidistant from their intersection, its direction must make equal supplementary/adjacent angles with the two lines. Thus $$\left|\frac{m-1/2}{1+m/2}\right|=\left|\frac{m+1}{1-m}\right|.$$ Squaring, $$\frac{(m-1/2)^2}{(1+m/2)^2}=\frac{(m+1)^2}{(1-m)^2}.$$ Multiply through: $$\frac{(2m-1)^2}{(m+2)^2}=\frac{(m+1)^2}{(1-m)^2}.$$ So $$(2m-1)^2(1-m)^2=(m+1)^2(m+2)^2.$$ Taking square roots gives two cases: ### Case 1: $$(2m-1)(1-m)=(m+1)(m+2)$$ $$-2m^2+3m-1=m^2+3m+2$$ $$-3m^2-3=0$$ $$m^2=-1$$ No real solution. ### Case 2: $$(2m-1)(1-m)=-(m+1)(m+2)$$ $$-2m^2+3m-1=-m^2-3m-2$$ $$-m^2+6m+1=0$$ $$m^2-6m-1=0.$$ Thus $$m=3\pm \sqrt{10}.$$ Therefore the sum of all possible distinct values is $$(3+\sqrt{10})+(3-\sqrt{10})=6.$$ --- 7. **Final answer** The sum of all possible distinct values of the slope is $$\boxed{6}.$$ So the correct option is **D**.

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Correct answer: D

Given equal sides

The two equal sides of the isosceles triangle lie along $L_1:",-x+2y=4 \quad \Rightarrow \quad y=\frac{x}{2}+2$ and $L_2:",x+y=4 \quad \Rightarrow \quad y=-x+4.$

These two lines meet at the vertex where the equal sides of the isosceles triangle intersect.

Find the vertex of intersection

Solve $-x+2y=4, \qquad x+y=4.$

From $x+y=4$ , we get $x=4-y$ . Substitute into the first equation: $-(4-y)+2y=4$ $-4+y+2y=4$ $3y=8 \Rightarrow y=\frac{8}{3}.$ Then $x=4-\frac{8}{3}=\frac{4}{3}.$

So the common vertex is $A\left(\frac{4}{3},\frac{8}{3}\right).$

Angle bisectors of the two equal sides

So first find the angle bisectors of the pair of lines.

Write lines in standard form: $L_1: -x+2y-4=0, \qquad L_2: x+y-4=0.$

Angle bisectors satisfy $\frac{-x+2y-4}{\sqrt{(-1)^2+2^2}}=\pm \frac{x+y-4}{\sqrt{1^2+1^2}}$ $\frac{-x+2y-4}{\sqrt5}=\pm \frac{x+y-4}{\sqrt2}.$

Find slopes of the angle bisectors

Instead of solving fully, use the fact that if a bisector has slope $s$ , then it makes equal angles with the two given lines.

The slopes of the given lines are $m_1=\frac12, \qquad m_2=-1.$

Case 1:

Case 2:

So the two angle bisectors have slopes $4+3\sqrt{10}, \qquad 4-3\sqrt{10}.$

Slope of the third side

The third side is perpendicular to an angle bisector, so if its slope is $m$ , then $m=-\frac1s.$

Hence possible slopes are

\qquad m_2=-\frac{1}{4-3\sqrt{10}}.$$ Now add them: $$m_1+m_2=-\left(\frac{1}{4+3\sqrt{10}}+\frac{1}{4-3\sqrt{10}}\right).$$ Using $$\frac{1}{a+b}+\frac{1}{a-b}=\frac{2a}{a^2-b^2},$$ with $a=4$, $b=3\sqrt{10}$, $$m_1+m_2=-\frac{2\cdot 4}{16-90} =-\frac{8}{-74}=\frac{4}{37}.$$ This looks inconsistent with the options, so let us use a cleaner geometric approach. --- 6. **Cleaner approach using angle formula** Let the apex angle between the equal sides be $\theta$. Then the base makes equal angles with the two sides. If the base has slope $m$, then the perpendicular from apex to base is an angle bisector. Let $\phi_1=\tan^{-1}(1/2)$ and $\phi_2=\tan^{-1}(-1)$ be inclinations of the equal sides. The bisector inclinations are $$\frac{\phi_1+\phi_2}{2}, \qquad \frac{\phi_1+\phi_2}{2}+90^\circ.$$ The base inclinations are each $90^\circ$ from these, i.e. again the pair of lines whose slopes satisfy the quadratic obtained from angle-bisector relation. A more direct algebraic route is to note: For a line of slope $m$ to cut the two given lines at points equidistant from their intersection, its direction must make equal supplementary/adjacent angles with the two lines. Thus $$\left|\frac{m-1/2}{1+m/2}\right|=\left|\frac{m+1}{1-m}\right|.$$ Squaring, $$\frac{(m-1/2)^2}{(1+m/2)^2}=\frac{(m+1)^2}{(1-m)^2}.$$ Multiply through: $$\frac{(2m-1)^2}{(m+2)^2}=\frac{(m+1)^2}{(1-m)^2}.$$ So $$(2m-1)^2(1-m)^2=(m+1)^2(m+2)^2.$$ Taking square roots gives two cases: ### Case 1: $$(2m-1)(1-m)=(m+1)(m+2)$$ $$-2m^2+3m-1=m^2+3m+2$$ $$-3m^2-3=0$$ $$m^2=-1$$ No real solution. ### Case 2: $$(2m-1)(1-m)=-(m+1)(m+2)$$ $$-2m^2+3m-1=-m^2-3m-2$$ $$-m^2+6m+1=0$$ $$m^2-6m-1=0.$$ Thus $$m=3\pm \sqrt{10}.$$ Therefore the sum of all possible distinct values is $$(3+\sqrt{10})+(3-\sqrt{10})=6.$$ --- 7. **Final answer** The sum of all possible distinct values of the slope is $$\boxed{6}.$$ So the correct option is **D**.

Straight Lines and Pair of Straight Lines question

Straight Lines and Pair of Straight Lines question

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