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Question

Mathematics Question on Straight lines

Suppose the point P(1,1)P(1,1) is translated to QQ in the direction of y = 2x. If PQ=1PQ = 1,then Qis

A

(2,0)(2, 0)

B

(0,2)(0,2)

C

(2+12,2+12)(\dfrac{√2+1}{√2},\dfrac{√2+1}{√2})

D

(5+15,5+25)(\dfrac{√5+1}{√5},\dfrac{√5+2}{√5})

E

(2+32,32)(\dfrac{2+√3}{2},\dfrac{3}{2})

Answer

(5+15,5+25)(\dfrac{√5+1}{√5},\dfrac{√5+2}{√5})

Explanation

Solution

Given that:

y=2x y=2x, is the equation of the line passing through the points P(1,1)

So , y1=2(x1)y−1=2(x−1)

y=2x1⇒y=2x−1

Let the position of QQ be (x1,y1)(x1​,y1​)

Since, Q(x1,y1)Q(x1​,y1​) lies on this line

y1=2x11y_1​=2x_1​−1 ----------(1)

Also, given P is translated to Q by a unit distance (As PQ=1)

PQ=1PQ=1

(x11)2+(y11)2=1⇒(x_1​−1)^2+(y_1​−1)^2=1

(x11)2+(2x12)2=1⇒(x_1​−1)^2+(2x_1​−2)^2=1 (from equation (1))

5x1210x1+4=0⇒5x_1^2​−10x_1​+4=0

Now to get values of x1x_1 from the quadratic equation we can write:

x1=10±20​​10⇒x_1​=\dfrac{10±√20​​}{10}

x1=5±5​​5⇒x_1​=\dfrac{5±√5​​}{5}
x1=1±15⇒x_1​=1±\dfrac{1​}{√5​}

Substitute the above value in equation(1), we get

y1=1±25⇒y_1​=1±\dfrac{2​}{√5​}
Hence, the new position of PP is (1±15,1±25)(Ans)(1±\dfrac{1​}{√5​}​,1±\dfrac{2​}{√5​}​) (\text{Ans})