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The equation to the sides of a triangle are $x-3y=0$, $4x+3y=5$ and $3x+y=0$. The line $3x-4y=0$passes through
Area of the parallelogram formed by the lines ${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0$,${{a}_{1}}x+{{b}_{1}}y+{{d}_{1}}=0$and ${{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0$, ${{a}_{2}}x+{{b}_{2}}y+{{d}_{2}}=0$is
Area of the parallelogram whose sides are $x\cos \alpha +y\sin \alpha =p$ $x\cos \alpha +y\sin \alpha =q,\,\,$ $x\cos \beta +y\sin \beta =r$ and $x\cos \beta +y\sin \beta =s$ is
The area of the triangle bounded by the straight line $ax+by+c=0,\,\,\,\,(a,b,c\ne 0)$ and the coordinate axes is
The triangle formed by the lines $x+y-4=0,\,$ $3x+y=4,$ $x+3y=4$ is
Two lines are drawn through (3, 4), each of which makes angle of $45^o$ with the line $x-y=2$, then area of the triangle formed by these lines is
The area of the triangle formed by the line $x\sin \alpha +y\cos \alpha =\sin 2\alpha $and the coordinates axes is
The area of a parallelogram formed by the lines $ax\pm by\pm c=0$, is
The triangle formed by ${{x}^{2}}-9{{y}^{2}}=0$and $x=4$is
A point moves so that square of its distance from the point (3, – 2) is numerically equal to its distance from the line $5x-12y=13$. The equation of the locus of the point is

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