Free Geometry Set I - 03 Practice Test - CAT 

$1^{st}$ method: - In a right angled triangle

Circumradius = $\frac{hypotenuse}{2}$

The coordinates of circumcentre = (3,1)

$2^{nd}$ method: - If O(x₁, y₁) is the circumcentre, then $O{A}^2=O{B}^2$.
$(x_1-2{)}^2+(Y_1-3{)}^2=(X_1-4{)}^2+(Y_1+1{)}^2$
$-4X_1-6Y_1+8X_1-2Y_1=17-13=4$
i.e., $4X_1-8Y_1=4$ or $X_1-2Y_1=1$ ..................(I)
$O{B}^2=O{C}^2$
$X_1-4{)}^2+(Y_1+1{)}^2=(X_1-4{)}^2+(Y_1-3{)}^2$
$2Y_1+6Y_1=8$
$8Y_1=8$ ....................(II)
$Y_1=1$
From (i) $X_1=3$
$X_1,Y_1)=(3,1)$

The simplest way to find this is to put a rectangle around the given triangle and then subtract off the areas of the right triangles that surround it. (See Figure) The area of the entire rectangle is 126, but after subtracting the areas of the 3 right triangles, whose areas are 54, 3, and 42, we are left with an area of 27.

Given lines are 5x + 12y + 18 = 0……………….(i) and 5x + 12y – 21 = 0………………….(ii)

Length of the side = distance between lines (i) and (ii) is given by

D = $|\frac{C_2-C_1}{\sqrt{a^2+b^2}}|=|\frac{-21-18}{\sqrt{25+144}}|=|\frac{-39}{13}|=3$

Length of the diagonal = $3\sqrt{2};$ Option(a).

$1^{st}$ method: - The simplest way to find this is to put a rectangle around the given quadrilateral and then subtract off the areas of the right triangles that surround it. (See Figure) The area of the entire rectangle is 40, but after subtracting the areas of the 4 right triangles, whose areas are $\frac{5}{2}$, 4, 9 and 3. we are left with an area of 21.5.

$2^{nd}$ method: - Area of quadrilateral = $2(x_1{y}_2-x_2{y}_1)+(x_2{y}_3-x_3{y}_2)+(x_3{y}_4-x_4{y}_3)+(x_4{y}_1-x_1{y}_4)$

=$\frac{1}{2}(-4\times 0-(-1)\times 2)+(-1\times 1-4\times 0)+(4\times 5-2\times 1)+(2\times 2-(-4)\times 5)=21.5$

2x + 2y + 1 = 0 …………….(i)

8x – 2y + 9 = 0 …………………………(ii)

After solving equation (i) and (ii)

x = - 1 and y = $\frac{1}{2}$

$(-1,\frac{1}{2})$; which is in $2^{nd}$ quadrant.

Option(a)

Slope of the line = $\frac{y_2-y_1}{x_2-x_1}=\frac{7-5}{9-7}=1$ 

Here $(x_1,y_1)$ = (7, 5) and $(x_2,y_2)$ = (9, 7)

Even without calculations it is clear that $(\sqrt{2},\frac{31}{3})$ cannot lie on the line y = 5x + 3. If x is irrational, y must also be irrational. Option(d)

Given, AB = 2BC

AB : BC = 2 : 1

Let C BE (h, k), then

$(\frac{2h\pm 1}{2\pm 1};\frac{2k\pm 7}{2\pm 1})$ = (6, - 3) [B may divide AC internally or externally]

2h + 1 = 18 and 2k + 7 = - 9

(or) 2h - 1 = 6 , 2k - 7 = - 3

2h = 17 and 2k = - 16 (or) 2h = 7, 2k = 4

h = $\frac{17}{2}$ and k = - 8 (or) h = $\frac{7}{2}$; k = 2

C = $(\frac{17}{2};-8)$ or $(\frac{17}{2};2)$

Option(b)

A(1, - 2) ,B(- 3, 0) and C(5, 6)

AB = $\sqrt{20}$, BC = $\sqrt{100}$ , AC = $\sqrt{80}$

$B{C}^2=A{B}^2+A{C}^2$

ABC is a right angled triangle with the right angle at A.

Circumradius is half the length of the hypotenuse.

R = $\frac{1}{2}\times$ BC, BC = 5. Option(a).

Area of pentagon = Area of triangle ABE + Area of rectangle BCDE

$\frac{1}{2}\times 4\times 2+5\times 4=24$

The given tangents are

2x+3y-9 = 0

4x+6y+19 =0

Or 2x + 3y + $\frac{19}{2}$ = 0

Which are parallel, than distance between parallel tangents must be diameter of the circle than diameter.
$\frac{|\frac{19}{2}-(-9)|}{\sqrt{2^2+3^2}}=\frac{37}{2\sqrt{13}}$

Radius = $\frac{1}{2}$ diameter = $\frac{37}{4\sqrt{13}}$

option (b)

To have an x-intercept 4, the value at y=0 should give x=4.

Put y=0 in all 3 answer options.

Option a and b give x=4

$1^{st}$ method: - equation of any straight line passing through the intersection of the lines x + 2y = 3 and 2x – 3y= 4 is

$\lambda$(x + 2y – 3) + (2x – 3y – 4) = 0

Since it passes through the point (2, 1)

$\lambda$(2 + 2 – 3) + (4 – 3 – 4) = 0

 $\lambda$ - 3 = 0

$\lambda$ = 3

Now substituting this value of $\lambda$  in (i), we get

3(x + 2y – 3) + (2x – 3y – 4) = 0

5x + 3y – 13 = 0

$2^{nd}$ method: - The straight line passing through the point (2, 1), put x = 2 and y = 1only option (a) and (b) will satisfy. Now, intersection point of the lines x + 2y = 3 and 2x – 3y= 4 is $(\frac{17}{7},\frac{2}{7}),$ Put x = $\frac{17}{7}$ and y = $\frac{2}{7}$ only option(a) will satisfy. Option(a).

The given circle is $x^2+y^2-2x-4y-20=0$ with centre (1, 2) and radius = $\sqrt{1+4+20}=5$.

And required circle has radius 5 hence circles touch each other externally.

Since point of contact is P(5, 5).

P is the mid point of $C_1$ and $C_2$, let co – ordinate of centre $C_2$ is (h, k) then

$\frac{1+h}{2}=5;h=9$

$\frac{2+k}{2}=5$; k = 8

And, Equation of required circle is $(x-9{)}^2+(y-8{)}^2=5^2$

$x^2+y^2-18x-16y+120=0;$ Option(a)

From the given figure, we can easily determine that the co-ordinates of the foot of the perpendicular from (2, 4) upon the line x + y = 4 are (1, 3)