CUET UG 2025 — Mathematics Algebra

The triangle has vertices $(-2, 4)$(−2,4), $(2, -6)$(2,−6), and $(k, 4)$(k,4) where $k > 0$k>0.

The area of the triangle is 35 square units.

For a triangle with vertices $(x_1, y_1)$(x1​,y1​), $(x_2, y_2)$(x2​,y2​), and $(x_3, y_3)$(x3​,y3​), the area is:

$$\text{Area} = \frac{1}{2}|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$$Area=21​∣x1​(y2​−y3​)+x2​(y3​−y1​)+x3​(y1​−y2​)∣

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Let $(x_1, y_1) = (-2, 4)$(x1​,y1​)=(−2,4), $(x_2, y_2) = (2, -6)$(x2​,y2​)=(2,−6), and $(x_3, y_3) = (k, 4)$(x3​,y3​)=(k,4).

Substituting into the formula:

$$35 = \frac{1}{2}|(-2)[(-6) - 4] + 2[4 - 4] + k[4 - (-6)]|$$35=21​∣(−2)[(−6)−4]+2[4−4]+k[4−(−6)]∣

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Calculating each term:

$$(-2)[(-6) - 4] = (-2)(-10) = 20$$(−2)[(−6)−4]=(−2)(−10)=20

$$2[4 - 4] = 2(0) = 0$$2[4−4]=2(0)=0

$$k[4 - (-6)] = k(10) = 10k$$k[4−(−6)]=k(10)=10k

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The equation becomes:

$$35 = \frac{1}{2}|20 + 0 + 10k|$$35=21​∣20+0+10k∣

$$35 = \frac{1}{2}|20 + 10k|$$35=21​∣20+10k∣

$$70 = |20 + 10k|$$70=∣20+10k∣

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The absolute value equation gives two cases:

Case 1: $20 + 10k = 70$20+10k=70

$$10k = 50$$10k=50

$$k = 5$$k=5

Case 2: $20 + 10k = -70$20+10k=−70

$$10k = -90$$10k=−90

$$k = -9$$k=−9

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Since $k > 0$k>0, the value $k = -9$k=−9 is rejected.

Therefore, $k = 5$k=5.