332354 + Interview Questions in Secondary School in Math

1.	Find the locus of the image of the point (2,3) on the line (2x-3y+4) + lambda (x-2y+3). Here lambda is any constant.
Answer» Given: A point P(2,3) A line (2x-3y+4)+λ(x-2y+3)=0 To FIND: The locus of the image of the point P(2,3) on the line (2x-3y+4)+λ(x-2y+3)=0 Diagram: $\setlength{\unitlength}{1cm}\thicklines\begin{picture}(30,15)\put(0,0){\line(2,1){5}}\put(0,2.5){\line(2,-1){5}}\thinlines\multiput(0,1.23)(0.4,0){14}{\line(1,0){0.2}}\multiput(2.49,2.95)(0,-0.4){9}{\line(0,-1){0.2}}\put(3.39,2.6){\circle{0.1}}\put(3.7,2.7){$P(2,3)$}\put(1.48,-0.24){\circle{0.1}}\put(0.5,-0.7){$R(h,k)$}\put(2.48,1.24){\circle{0.2}}\put(1.6,1.7){$Q(1,2)$}\put(1.48,-0.24){\line(2,3){2}}\put(6.5,2){$(2x-3y+4=0)$}\put(6.67,0.4){$x-2y+3=0$}\put(6.5,2.1){\vector(-1,0){2}}\put(6.6,0.5){\vector(-1,0){2.1}}\end{picture}$ Here, dotted lines and line PQR are some of those lines which passes through point Q. Solution:* Here, the equation (2x-3y+4)+λ(x-2y+3)=0 represents the equations of all those lines which passes through the intersection of line 2x-3y+4=0 and x-2y+3=0 LET the intersection point of lines 2x-3y+4=0 and x-2y+3=0 be Q Now, here $\rm{2x-3y+4=0}------(1)$ $\rm{x-2y+3=0}-------(2)$ On multiplying (2) by 2, we get $\rm{2x-4y+6=0}-------(3)$ On subtracting (2) from (1), we get $\rm{2x-3y+4-(2x-4y+6)=0}$ $\rm{2x-3y+4-2x+4y-6=0}$ $\rm{y-2=0}$ $\rm{y=2}$ On putting y=2 in (1), we get $\rm{2x-3(2)+4=0}$ $\rm{2x-6+4=0}$ $\rm{2x-2=0}$ $\rm{2x=2}$ $\rm{x=\dfrac{2}{2}}$ $\rm{x=1}$ So, the coordinates of point Q is (1,2) $\rule{190}{1}$ Now, let the image of point P be R(h,k) Since, point R is image of P, So,by geometry $\longrightarrow\rm{PQ=QR}$ Using Distance formula, $\sf{\sqrt{(1-2)^{2}+(2-3)^{2}}=\sqrt{(h-1)^{2}+(k-2)^{2}}}$ On SQUARING both the sides, we get $\rm{(1-2)^{2}+(2-3)^{2}=(h-1)^{2}+(k-2)^{2}}$ $\sf{(-1)^{2}+(-1)^{2}=h^{2}+(1)^{2}-2h+k^{2}+(2)^{2}-4k}$ $\rm{1+1=h^{2}+1-2h+k^{2}+4-4k}$ $\rm{2=h^{2}-2h+k^{2}-4k+5}$ $\rm{h^{2}-2h+k^{2}-4k+5=2}$ $\rm{h^{2}+k^{2}-2h-4k+5-2=0}$ $\rm{h^{2}+k^{2}-2h-4k+3=0}$ On substituting (h,k) with (x,y), we get $\longrightarrow\rm\green{x^{2}+y^{2}-2x-4y+3=0}$ Hence, the locus of the image of the point P(2,3) on the line (2x-3y+4)+λ(x-2y+3)=0 is x²+y²-2x-4y+3=0.

1.

Find the locus of the image of the point (2,3) on the line (2x-3y+4) + lambda (x-2y+3). Here lambda is any constant.

Answer»

Given:

A point P(2,3)

A line (2x-3y+4)+λ(x-2y+3)=0

To FIND:

The locus of the image of the point P(2,3) on the line (2x-3y+4)+λ(x-2y+3)=0

Diagram:

$\setlength{\unitlength}{1cm}\thicklines\begin{picture}(30,15)\put(0,0){\line(2,1){5}}\put(0,2.5){\line(2,-1){5}}\thinlines\multiput(0,1.23)(0.4,0){14}{\line(1,0){0.2}}\multiput(2.49,2.95)(0,-0.4){9}{\line(0,-1){0.2}}\put(3.39,2.6){\circle*{0.1}}\put(3.7,2.7){$P(2,3)$}\put(1.48,-0.24){\circle*{0.1}}\put(0.5,-0.7){$R(h,k)$}\put(2.48,1.24){\circle*{0.2}}\put(1.6,1.7){$Q(1,2)$}\put(1.48,-0.24){\line(2,3){2}}\put(6.5,2){$(2x-3y+4=0)$}\put(6.67,0.4){$x-2y+3=0$}\put(6.5,2.1){\vector(-1,0){2}}\put(6.6,0.5){\vector(-1,0){2.1}}\end{picture}$

Here, dotted lines and line PQR are some of those lines which passes through point Q.

Solution:

Here, the equation (2x-3y+4)+λ(x-2y+3)=0 represents the equations of all those lines which passes through the intersection of line 2x-3y+4=0 and x-2y+3=0

LET the intersection point of lines 2x-3y+4=0 and x-2y+3=0 be Q

Now, here

$\rm{2x-3y+4=0}------(1)$

$\rm{x-2y+3=0}-------(2)$

On multiplying (2) by 2, we get

$\rm{2x-4y+6=0}-------(3)$

On subtracting (2) from (1), we get

$\rm{2x-3y+4-(2x-4y+6)=0}$

$\rm{2x-3y+4-2x+4y-6=0}$

$\rm{y-2=0}$

$\rm{y=2}$

On putting y=2 in (1), we get

$\rm{2x-3(2)+4=0}$

$\rm{2x-6+4=0}$

$\rm{2x-2=0}$

$\rm{2x=2}$

$\rm{x=\dfrac{2}{2}}$

$\rm{x=1}$

So, the coordinates of point Q is (1,2)

$\rule{190}{1}$

Now, let the image of point P be R(h,k)

Since, point R is image of P,

So,by geometry

$\longrightarrow\rm{PQ=QR}$

Using Distance formula,

$\sf{\sqrt{(1-2)^{2}+(2-3)^{2}}=\sqrt{(h-1)^{2}+(k-2)^{2}}}$

On SQUARING both the sides, we get

$\rm{(1-2)^{2}+(2-3)^{2}=(h-1)^{2}+(k-2)^{2}}$

$\sf{(-1)^{2}+(-1)^{2}=h^{2}+(1)^{2}-2h+k^{2}+(2)^{2}-4k}$

$\rm{1+1=h^{2}+1-2h+k^{2}+4-4k}$

$\rm{2=h^{2}-2h+k^{2}-4k+5}$

$\rm{h^{2}-2h+k^{2}-4k+5=2}$

$\rm{h^{2}+k^{2}-2h-4k+5-2=0}$

$\rm{h^{2}+k^{2}-2h-4k+3=0}$

On substituting (h,k) with (x,y), we get

$\longrightarrow\rm\green{x^{2}+y^{2}-2x-4y+3=0}$

Hence, the locus of the image of the point P(2,3) on the line (2x-3y+4)+λ(x-2y+3)=0 is x²+y²-2x-4y+3=0.

3.	A ratio equivalent to 3:7 is:
Answer» 3:7=6:14 or 9:21 Hope it HELPS you..

6.	A trader Mark's his good at 50 percent above the cost price but allows a discount of 20 percent on the market price. Calculate his gain percent.
Answer» 50 percent of cost price 20 percent discount 50-20 = 30 percent as gain percent it also can be more than 30 but here no cost price or SELLING price is GIVEN So gain or profit percent is 30 percent

7.	(2/5)+(8/9)= pls solve to
Answer» The REQUIRED ANSWER is 1.2. 2/5=0.4 8/9=0.8 Therefore, 0.4+0.8=1.2 Hope it HELPS you..

16.	If we stand in front on the sun at 3 am .which angle is formed
Answer» 75° @ngle is FORMED..........

35.	Pls tell answer correctly
Answer» We know that, a³ + b³ + c³ -3ABC = (a + b + c )(a² + b² + c² -ab -bc-ca) now , a + b + c = 5 ab + bc + ca = 10 (a + b + c)² = a² + b² + c² +2(ab + bc+ca) (5)² -2×10 = a² + b² + c² a² + b² + c² =5 hence , a³ + b³ +c³ -3abc = ( a + b + c )(a² + b² + c² -ab- bc-ca) =( 5)( 5 - 10) = 5 × (-5) = -25 hence proved// Hope it HELPS you...

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