Opposite angular points of a square are (3,4) and (1, - 1) . Then co-ordinates of other two points a

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9.Straight Line

medium

The opposite angular points of a square are $(3,\;4)$ and $(1,\; - \;1)$. Then the co-ordinates of other two points are

$D\,\left( {\frac{1}{2},\,\,\frac{9}{2}} \right)\,,\,\,B\,\left( { - \frac{1}{2},\,\,\frac{5}{2}} \right)$

$D\,\left( {\frac{1}{2},\,\,\frac{9}{2}} \right)\,,\,\,B\,\left( {\frac{1}{2},\,\,\frac{5}{2}} \right)$

$D\,\left( {\frac{9}{2},\,\,\frac{1}{2}} \right)\,,\,\,B\,\left( { - \frac{1}{2},\,\,\frac{5}{2}} \right)$

None of these

Solution

Let m be the slope of a line inclined at an angle of ${45^o}$ to $AC$, then $\tan {45^o} = \pm \frac{{m – \frac{5}{2}}}{{1 + m.\frac{5}{2}}} \Rightarrow m = – \frac{7}{3},\frac{3}{7}$.

Thus, let the slope of $AB$ or $DC$ be $3/7$ and that of $AD$ or $BC$ be $ – \frac{7}{3}$ . Then equation of $AB$ is $3x – 7y + 19 = 0$.

Also the equation of $BC$ is $7x + 3y – 4 = 0$.

On solving these equations, we get, $B\,\,\,\left( { – \frac{1}{2},\frac{5}{2}} \right)$.

Now let the coordinates of the vertex $D$ be $(h, k)$. Since the middle points of $AC$ and $BD$ are same,

therefore $\frac{1}{2}\left( {h – \frac{1}{2}} \right)\, = \frac{1}{2}(3 + 1) \Rightarrow h = \frac{9}{2}$, $\frac{1}{2}\left( {k + \frac{5}{2}} \right) = \frac{1}{2}(4 – 1)$

==> $k = \frac{1}{2}$. Hence, $D = \left( {\frac{9}{2},\,\frac{1}{2}} \right)$.

Standard 11

Mathematics

Let $A B C$ and $A B C^{\prime}$ be two non-congruent triangles with sides $A B=4$, $A C=A C^{\prime}=2 \sqrt{2}$ and angle $B=30^{\circ}$. The absolute value of the difference between the areas of these triangles is

hard

(IIT-2009)

If the equation of base of an equilateral triangle is $2x – y = 1$ and the vertex is $(-1, 2)$, then the length of the side of the triangle is

medium

The co-ordinates of the vertices $A$ and $B$ of an isosceles triangle $ABC (AC = BC)$ are $(-2,3)$ and $(2,0)$ respectively. $A$ line parallel to $AB$ and having a $y$ -intercept equal to $\frac{43}{12}$ passes through $C$, then the co-ordinates of $C$ are :-

normal

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easy

(IIT-1983)

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