a) Find the equation of a straight line through the mid point of the line segment connecting (2, 4) and (2, 4) and parallel to the line 3x 2y = 4 . [2] b) Find the equation of a circle having radius…

a) Equation of a straight line through the midpoint and parallel to a given line

Step 1: Find the midpoint of the line segment connecting $(2,-4)$ and $(2,4)$.

The midpoint $M$ of a line segment joining two points $(x_1,y_1)$ and $(x_2,y_2)$ is given by: $$M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$$

Substituting the given points $(2,-4)$ and $(2,4)$: $$M=\left(\frac{2+2}{2},\frac{-4+4}{2}\right)=(2,0)$$

Step 2: Find the slope of the given line $3x-2y=4$.

Rewrite the equation in slope-intercept form $y=mx+c$: $$3x-2y=4⟹-2y=-3x+4⟹y=\frac{3}{2}x-2$$ The slope $m$ of the given line is $\frac{3}{2}$.

Step 3: Write the equation of the line parallel to the given line and passing through the midpoint $(2,0)$.

Since parallel lines have the same slope, the slope of the desired line is also $\frac{3}{2}$. Using the point-slope form: $$y-y_1=m(x-x_1)$$ Substituting $(x_1,y_1)=(2,0)$ and $m=\frac{3}{2}$: $$y-0=\frac{3}{2}(x-2)⟹y=\frac{3}{2}x-3$$

Convert to standard form: $$2y=3x-6⟹3x-2y-6=0$$

Final Answer: The equation of the line is $3x-2y-6=0$.

---

b) Equation of a circle with given radius and diameters

Step 1: Find the center of the circle.

The center of the circle is the point of intersection of the two given diameters: $$x+2y=8\text{(1)}$$ $$x+y=6\text{(2)}$$

Solve equations (1) and (2) simultaneously: From (2): $x=6-y$.

Substitute into (1): $$(6-y)+2y=8⟹6+y=8⟹y=2$$ Then, $x=6-2=4$.

So, the center of the circle is $(4,2)$.

Step 2: Write the equation of the circle.

The standard equation of a circle with center $(h,k)$ and radius $r$ is: $$(x-h{)}^2+(y-k{)}^2=r^2$$

Given $r=10$, $h=4$, and $k=2$: $$(x-4{)}^2+(y-2{)}^2=10^2$$ $$(x-4{)}^2+(y-2{)}^2=100$$

Final Answer: The equation of the circle is $(x-4{)}^2+(y-2{)}^2=100$.

---

c) Evaluate: ${\lim }_{x\to p}\frac{x^2-p^2}{\tan (x-p)}$

Step 1: Direct substitution check

Substitute $x=p$: $$\frac{p^2-p^2}{\tan (p-p)}=\frac{0}{\tan (0)}=\frac{0}{0}$$ This is an indeterminate form, so apply L'Hôpital's Rule.

Step 2: Apply L'Hôpital's Rule

Differentiate the numerator and denominator with respect to $x$: $$\text{Numerator: }\frac{d}{dx}(x^2-p^2)=2x$$ $$\text{Denominator: }\frac{d}{dx}(\tan (x-p))={\sec }^2(x-p)$$

Now, the limit becomes: $$\underset{x\to p}{\lim }\frac{2x}{{\sec }^2(x-p)}$$

Step 3: Simplify and evaluate

Substitute $x=p$: $$\frac{2p}{{\sec }^2(0)}=\frac{2p}{1}=2p$$

Alternatively, use trigonometric identities and Taylor series expansion for $\tan (x-p)$ around $x=p$: $$\tan (x-p)\approx (x-p)+\frac{(x-p{)}^3}{3}+\cdots$$ For small $h=x-p$, $\tan (h)\approx h$, so: $$\frac{x^2-p^2}{\tan (x-p)}=\frac{(x-p)(x+p)}{x-p}=x+p\text{(for }x\ne p\text{)}$$ Taking the limit as $x\to p$: $$\underset{x\to p}{\lim }(x+p)=2p$$

Final Answer: The value of the limit is $2p$.