Solve the following questions.......don't spam ......❌⭕❌❗❗�

1.	Solve the following questions.......don't spam ......❌⭕❌❗❗
Answer» 5. If two lines intersect each other, then they are not parallel. Therefore their slopes should not be same. The slope of the line $4x+py+8=0$ is, $\longrightarrow m_1=-\dfrac{<klux>4</klux>}{p}$ And the slope of the line $2x+2y+2=0$ is, $\longrightarrow m_2=-\dfrac{2}{2}$ $\longrightarrow m_2=-1$ For the two lines intersect each other, $\longrightarrow m_1\neq m_2$ $\longrightarrow -\dfrac{4}{p}\neq-1$ $\longrightarrow \underline{\underline{p\neq4}}$ Hence, $\longrightarrow \underline{\underline{p\in\mathbb{R}-\{4\}}}$ 6. For the two lines being coincident, the two equations should be the same. Given equations are, $2\alpha x+(\alpha+\beta)y=28\quad\quad\dots(1)$ $2x+3y=7\quad\quad\dots(2)$ Multiplying (2) by 4, $8x+12y=28\quad\quad\dots(<klux>3</klux>)$ And on comparing LHS of (1) and (3), since RHS are same, $\longrightarrow 2\alpha=8$ $\longrightarrow\underline{\underline{\alpha=4}}$ And, $\longrightarrow \alpha+\beta=12$ $\longrightarrow 4+\beta=12$ $\longrightarrow \underline{\underline{\beta=8}}$ 7. (i) For a unique solution, the lines represented by these two equations should not be parallel, hence their slopes should not be same. The slope of the line $x+2y-5=0$ is, $\longrightarrow m_1=-\dfrac{1}{2}$ And that of $3x+ky-15=0$ is, $\longrightarrow m_2=-\dfrac{3}{k}$ Then we have, $\longrightarrow m_1\neq m_2$ $\longrightarrow -\dfrac{1}{2}\neq-\dfrac{3}{k}$ $\longrightarrow \underline{\underline{k\neq6}}$ Therefore, $\longrightarrow\underline{\underline{k\in\mathbb{R}-\{6\}}}$ 7. (ii) For $k=6,$ the system would have infinite no. of solutions because the lines BECOME coincident. Therefore there are no values for k, for the system having no solution. 8. (i) For no solutions, the lines represented by these two lines should be parallel to each other, hence their slopes should be same. The slope of the line $px+3y-3=0$ is, $\longrightarrow m_1=-\dfrac{p}{3}$ And that of the line $12x+py-6=0$ is, $\longrightarrow m_2=-\dfrac{12}{p}$ For the two being parallel, $\longrightarrow m_1=m_2$ $\longrightarrow -\dfrac{p}{3}=-\dfrac{12}{p}$ $\Longrightarrow\underline{\underline{p=-6}}$ The reason for $p\neq6$ here is given below. 8. (ii) For infinite solutions, the lines should be coincident. Hence the equations should be the same. The equations are, $px+3y=3\quad\quad\dots(1)$ $12x+py=6\quad\quad\dots(2)$ Multiplying (1) by 2, $2px+6y=6\quad\quad\dots(3)$ And on comparing (2) and (3), we get, $\longrightarrow\underline{\underline{p=6}}$ I.e., the system has no solutions only if $p=-6.$ And the system has infinite solutions only if $p=6.$

Solve the following questions.......don't spam ......❌⭕❌❗❗

Answer»

5. If two lines intersect each other, then they are not parallel. Therefore their slopes should not be same.

The slope of the line $4x+py+8=0$ is,

$\longrightarrow m_1=-\dfrac{<klux>4</klux>}{p}$

And the slope of the line $2x+2y+2=0$ is,

$\longrightarrow m_2=-\dfrac{2}{2}$

$\longrightarrow m_2=-1$

For the two lines intersect each other,

$\longrightarrow m_1\neq m_2$

$\longrightarrow -\dfrac{4}{p}\neq-1$

$\longrightarrow \underline{\underline{p\neq4}}$

Hence,

$\longrightarrow \underline{\underline{p\in\mathbb{R}-\{4\}}}$

6. For the two lines being coincident, the two equations should be the same.

Given equations are,

$2\alpha x+(\alpha+\beta)y=28\quad\quad\dots(1)$

$2x+3y=7\quad\quad\dots(2)$

Multiplying (2) by 4,

$8x+12y=28\quad\quad\dots(<klux>3</klux>)$

And on comparing LHS of (1) and (3), since RHS are same,

$\longrightarrow 2\alpha=8$

$\longrightarrow\underline{\underline{\alpha=4}}$

And,

$\longrightarrow \alpha+\beta=12$

$\longrightarrow 4+\beta=12$

$\longrightarrow \underline{\underline{\beta=8}}$

7. (i) For a unique solution, the lines represented by these two equations should not be parallel, hence their slopes should not be same.

The slope of the line $x+2y-5=0$ is,

$\longrightarrow m_1=-\dfrac{1}{2}$

And that of $3x+ky-15=0$ is,

$\longrightarrow m_2=-\dfrac{3}{k}$

Then we have,

$\longrightarrow m_1\neq m_2$

$\longrightarrow -\dfrac{1}{2}\neq-\dfrac{3}{k}$

$\longrightarrow \underline{\underline{k\neq6}}$

Therefore,

$\longrightarrow\underline{\underline{k\in\mathbb{R}-\{6\}}}$

7. (ii) For $k=6,$ the system would have infinite no. of solutions because the lines BECOME coincident. Therefore there are no values for k, for the system having no solution.