Question

(-4,-1), (-1.-1), (-1, 2) and (1, m) are the vertices of a square. Find I, m . pls send answer with steps

Answer 1

Answer:

Suppose the line joining the points A (3,4) & C (1,-1) be y=mx+c

Then, m=(4+1)/(3-2)=5/2

and c=y-mx= -1 - (5/2)*1= -7/2

So, the equation is y=(5/2)x - 7/2

Now, the other diagonal BD is perpendicular to AC at its midpoint E.

It is clear to see that E=((3+1)/2, (4+(-1))/2) = (2, 3/2) suppose the equation of BD be y=m’x+c’

Since E lies on BD, it will satisfy its equation.

So, m’=-1/m (since BD is perpendicular to AC)

=> m’=-1/5/2= -2/5

Also, c=y-m’x= 3/2-(-2/5)*2 = 23/10

So, eqn is y=(-2/5)x + 23/10

Now clearly, BE=AC/2.

=> sqrt ( (x–2)^2 + (y–3/2)^2) = 1/2 * sqrt( (3–1)^2 + (4-(-1))^2 )

=> (x–2)^2 + (y–3/2)^2 = 29/4

But since (x,y) lies on BD, so y=(-2/5)x + 23/10

=> (x–2)^2 + ( (-2/5)x + 23/10 -3/2)^2 = 29/4

=> (x–2)^2 + 4/25 * (x–2)^2 = 29/4

=> (x–2)^2 * 29/25 = 29/4

=> (x–2)^2 = 25/4

=> x = 2+-5/2

=> x = 9/2, -1/2

from y = (-2/5)x + 23/10, we have:

y = 1/2, 5/2. Ans

Answer 2

Answer:

$\huge{\underline{\underline{\mathbb{\red{Answer:-}}}}}$

Let the vertical are

M (1, 7), N (4, 2)

R (-1, 1) , Q (-4, 4)

$mn = \sqrt{(1 - {4}^{2} + (7 - {2)}^{2} } = \sqrt{9 + 25 } = \sqrt{34}$

$nr = \sqrt{(4 + {1)}^{2} + (2 + {1)}^{2} } = \sqrt{25 + 9} = \sqrt{34}$

$rq = \sqrt{( - 1 + {4)}^{2} + ( - 1 - {4)}^{2} } = \sqrt{34}$

$mq = \sqrt{1 + {4)}^{2} + (7 - {4)}^{2} } = \sqrt{34}$

All sides are equal, Hence MNRQ is a square P (2, P) is midpoint of

A ( 6,5 ) , B (-2, 11)

$p \: = \frac{ - 5 + 11}{2}$

$p = > 3$

Hope it's help you.......