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  • The ends of the base of an isosceles triangle are at and <img alt="(0,1)" src="https://entrancecorner.oncodecog

The ends of the base of an isosceles triangle are at (2,0) and (0,1) and the equation of one side is x=2 then the orthocentre of the triangle is

Option: 1

\mathrm{\left(\frac{3}{4}, \frac{3}{2}\right)}


Option: 2

\left(\frac{5}{4}, 1\right)


Option: 3

\left(\frac{3}{4}, 1\right)


Option: 4

\left(\frac{4}{3}, \frac{7}{12}\right)


Answers (1)

best_answer

2^{2}+\left(\mathrm{x}_{1}-1\right)^{2}=\mathrm{x}_{1}^{2}
+\mathrm{x}_{1}^{2}+1-2 \mathrm{x}_{1}=\mathrm{x}_{1}^{2}
5=2 \mathrm{x}_{1} \text { or } \mathrm{x}_{1}=5 / 2

Equation of (1) from (2,5 / 2) to the given base

y-5 / 2=2(x-2)
2 y-5=4(x-2)

at y=1 \mathrm{-3 / 4=x-2\, or \, x=5 / 4}\Rightarrow B

Posted by

himanshu.meshram

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