Question

Find the values of rr for which the line (r+5)x–(r–1)y+r2+4r−5=0(r+5)x–(r–1)y+r2+4r−5=0 is


1.parallel to the xx-axis

2.parallel to the line 2x−y+4=02x−y+4=0,

3.parallel to the yy-axis

4.perpendicular to the line x−y+32=0x−y+32=0


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Answer 1

Answer:

The correct answers are  1) r = -5 ; 2) r = 7 ;  3) r = 1 ; 4) r = -2

Step-by-step explanation:

1) Equation parallel to x-axis is y = 0

2) Parallel lines have same slope i.e. m1 = m2

3) Equation parallel to y-axis is x = 0

4) Two lines are perpendicular then slope m1 x m2 = -1

Answer 2

Answer:

The values of $r$ for given equation $(r + 5)x - (r - 1)y + r^{2} + 4r - 5 = 0$ , for given four conditions are $-5$, $7$, $1$ and $-2$ respectively.

Step-by-step explanation:

Given equation,

$(r + 5)x - (r - 1)y + r^{2} + 4r - 5 = 0$

The standard equation of line in terms of slope and point is given by, $y = mx + c$.

Where, $m$ being the slope of the equation.

• Converting the given equation to standard equation:

That is, $(r + 5)x + r^{2} + 4r - 5 = (r - 1)y$

             $\frac{1}{r - 1} ((r + 5)x + r^{2} + 4r - 5) = y$

             $y = (\frac{r + 5}{r - 1}) x + (\frac{r^{2} + 4r - 5}{r-1} )$

Thus, slope of the equation, $m = \frac{r + 5}{r - 1}$

• Condition $(1)$ : Parallel to X-X axis

The slope of any line parallel to X axis is zero. That is, $m = 0$.

Therefore,  $\frac{r + 5}{r - 1} = 0$

                   $r + 5 = 0$

                   $r = -5$.

• Condition $(2)$ : Parallel to line $2x - y + 4 = 0$

That is, $2x + 4 = y$

            $y = 2x + 4$ , where slope of equation is $2$.

When two lines are parallel, they have same slope; i.e., here $m = 2$.

Therefore, $\frac{r + 5}{r - 1} = 2$

               $r + 5 = 2(r - 1)$

               $r + 5 = 2r - 2$

                     $r = 7$.

• Condition $(3)$ : Parallel to Y-Y axis

The slope of any line parallel to Y axis is infinity. That is, $m =$ ∞.

Therefore,  $\frac{r + 5}{r - 1} =$ ∞

                   $r - 1 = 0$

                   $r = 1$.

• Condition $(4)$ : Perpendicular to line $x - y + 32 = 0$

That is, $y = x + 32$., where slope of equation is $1$.

When two lines are perpendicular, then the product of their slope is $- 1$, i.e., $m(1) = -1$

and thus slope of equation is, $m = -1$.

Therefore, $\frac{r + 5}{r - 1} = -1$

                 $r + 5 = -1(r - 1)$

                 $r + 5 = -r + 1$

                 $2r = -4$

                 $r = -2$.

Thus, the values of $r$ for all the above four conditions are $-5$, $7$, $1$ and $-2$.