factorize the following expressions :pq – prq
given (pq – prq) = (pq×1 – pq×r)taking pq common,(pq – prq) = pq(1 – r)
find and correct the errors in the following mathematical statements

q19) dear student

kindly refer to the link for the similar query :


https://www meritnation
com/ask-answer/question/q-23-and-24-q23-give-the-formula-of-each-of-the-follow/the-p-block-elements/11940305

regards using the long division method: [image] nirgrahamuk:

observe({

fv <- req(input$filter\_var)

td <- req(tabledata())

updateselectizeinput(inputid = "filter\_vals",

choices = pull(td,fv))

})

observe({

fv <- as.character(req(input$filter\_var))

td <- req(table… hint: assume the co-ordinates of $r$ as $(x,y)$ and as it is given that $p,q,r$ are collinear points. apply distance condition of collinear points which is if $p,q,r$ are three collinear points then they must satisfy $pq+qr=pr$ and get one equation. apply distance formula which states that if $a({{x}\_{1}},{{y}\_{1}})$ and $b({{x}\_{2}},{{y}\_{2}})$ are two points then the distance $ab=\sqrt{{{({{x}\_{2}}-{{x}\_{1}})}^{2}}+{{({{y}\_{2}}-{{y}\_{1}})}^{2}}}$ and get the second equation by using $pr=10$ , solve these two equations and get the co-ordinates of $r$. complete step-by-step solution:it is given in the question that $p, q, r$ are collinear points.collinear means lying on the same line which means that $p, q, r$ lies on the same straight line.let us suppose that the co-ordinates of $r$ is $(x,y)$. now, we have $p(7,7)$, $q(3,4)$, $r(x,y)$ and also $pr=10$.we have to use distance formula in order to find $pq,qr,pr$. so, according to distance formula if $a({{x}\_{1}},{{y}\_{1}})$ and $b({{x}\_{2}},{{y}\_{2}})$ then $ab=\sqrt{{{({{x}\_{2}}-{{x}\_{1}})}^{2}}+{{({{y}\_{2}}-{{y}\_{1}})}^{2}}}$.now, $p,q,r$ are three collinear points then they must satisfy,$\therefore pq+qr=pr$using distance formula, we get$\rightarrow \sqrt{{{(7-3)}^{2}}+{{(7-4)}^{2}}}+\sqrt{{{(3-x)}^{2}}+{{(4-y)}^{2}}}=10$$\rightarrow \sqrt{16+9}+\sqrt{{{(3-x)}^{2}}+{{(4-y)}^{2}}}=10$$\rightarrow 5+\sqrt{{{(3-x)}^{2}}+{{(4-y)}^{2}}}=10$\[\rightarrow \sqrt{{{(3-x)}^{2}}+{{(4-y)}^{2}}}=5\]squaring both sides, we get$\rightarrow {{(3-x)}^{2}}+{{(4-y)}^{2}}=25$$\rightarrow {{x}^{2}}+{{y}^{2}}-6x-8y+25=25$cancelling 25 from both sides, we get$\rightarrow {{x}^{2}}+{{y}^{2}}=6x+8y...................(1)$also, it is given that $pr=10$, using this we get$\therefore pr=10$\[\rightarrow \sqrt{{{(7-x)}^{2}}+{{(7-y)}^{2}}}=10\]squaring both sides, we get\[\rightarrow {{(7-x)}^{2}}+{{(7-y)}^{2}}=100\]$\rightarrow {{x}^{2}}+{{y}^{2}}-14x-14y+98=100$putting the value of ${{x}^{2}}+{{y}^{2}}$ from equation (1), we get$\rightarrow 6x+8y-14x-14y=100-98$$\rightarrow -8x-6y=2$now, expressing $y$ in terms of $x$, we get$\rightarrow y=\dfrac{-(2+8x)}{6}=\dfrac{-(1+4x)}{3}..................(2)$putting the value of $y$ from equation (2) in equation (1), we get$\rightarrow {{x}^{2}}+{{\left( \dfrac{-(1+4x)}{3} \right)}^{2}}=6x+\dfrac{-(1+4x)}{3}\times 8$$\rightarrow \dfrac{9{{x}^{2}}+1+8x+16{{x}^{2}}}{9}=\dfrac{18x-8-32x}{3}$$\rightarrow 25{{x}^{2}}+8x+1=3(-14x-8)$$\rightarrow 25{{x}^{2}}+8x+1+42x+24=0$$\rightarrow 25{{x}^{2}}+50x+25=0$$\rightarrow 25({{x}^{2}}+2x+1)=0$$\rightarrow {{(x+1)}^{2}}=0$$\rightarrow x+1=0\rightarrow x=-1$now, putting the value of $x$ in equation (2), we get$\rightarrow y=\dfrac{-(1+4\times -1)}{3}$$\rightarrow y=\dfrac{3}{3}=1$so, $x=-1$ and $y=1$.hence the co-ordinates of $r$ is $(-1,1)$. note: this question is simple and there are many other alternative ways to solve it but the tricky part is to choose which method to solve the question. there are three different conditions of collinearity as there is slope condition, distance condition, and area condition. as in the other two conditions distances are not involved, only the distance condition involves the use of distances. in the question, distance is mentioned so we choose that method to solve it. hence, students should carefully read the question and choose which method is suitable to solve the question according to the values given. several ways to do this, here's one:

to go from (-6,4) to (-2,5) you need to go 4 in the x direction and 1 in the y direction. if you move again 4 in the x direction and 1 in the y direction you end up at (2, 6). that's p2 solution: ` x = {:((-4/5,-1),(2/5,1)):}` hint: mathematics includes the study of topics which are related to quantity, structure, space and change. to solve this problem, we first try to simplify the expression. for doing so we let ${{x}^{2}}-4x=t$ and solve by using this consideration. in this way we can easily factorise the expression.complete step-by-step answer:according to a question, we are given the expression $\left( {{x}^{2}}-4x \right)\left( {{x}^{2}}-4x-1 \right)-20$. to simplify this complex expression, we let ${{x}^{2}}-4x=t$.this can be mathematically expressed as: $t(t-1)-20$.since, these types of factorization are familiar to us, so we try to factorise the equation in terms of t:${{t}^{2}}-t-20$now, by using the middle term splitting method of separating the variables, we get$\begin{align}  & {{t}^{2}}+4t-5t-20=0 \\  & t(t+4)-5(t+4)=0 \\  & (t-5)(t+4)=0 \\ \end{align}$ so, the factors of the expression in terms of t are (t-5) and (t+4).now, finally replacing t as ${{x}^{2}}-4x$,we get$\left( {{x}^{2}}-4x-5 \right)\left( {{x}^{2}}-4x+4 \right)=0$ on further simplifying this expression, by using middle term splitting method again in part one:$\begin{align}  & {{x}^{2}}-4x-5={{x}^{2}}+x-5x-5 \\  & \rightarrow x(x+1)-5(x+1)=0 \\  & (x-5)(x+1)=0 \\ \end{align}$ now, for the second part, using the identity of ${{(a-b)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$, we get${{x}^{2}}-4x+4={{(x-2)}^{2}}$$\therefore (x-5)(x+1){{(x-2)}^{2}}=0$ hence, the factors of the required equation as per problem statement are: $(x-5)(x+1){{(x-2)}^{2}}=0$ note: this question can be alternatively solved by expanding the powers of x in the given expression and then letting ${{x}^{2}}$ as another variable. but the complexity of the problem would increase by doing so. hence, for simplification we must assume some other variable which can be finally replaced in the original expression to obtain the answer. pbr3 Using [reflection](/questions/tagged/reflection "show questions tagged 'reflection'") and [expression-trees](/questions/tagged/expression-trees "show questions tagged 'expression-trees'") you can provide the parameters and then call `OrderBy` function, Instead of returning `Expression<Func<Task, T>>` and then calling `OrderBy`.


Note that `OrderBy` is an extension method and has implemented in both `System.Linq.Enumarable` and `System.Linq.Queryable` classes. The first one is for [linq-to-objects](/questions/tagged/linq-to-objects "show questions tagged 'linq-to-objects'") and the latter is for [linq-to-entities](/questions/tagged/linq-to-entities "show questions tagged 'linq-to-entities'"). [entity-framework](/questions/tagged/entity-framework "show questions tagged 'entity-framework'") needs the expression tree of the query in order to translate it to SQL commands. So we use the `Queryable` implementation.


It can be done by an extension method(explanations added as comments):



```
public static IOrderedQueryable<TSource> OrderBy<TSource>(
       this IQueryable<TSource> query, string propertyName)
{
    var entityType = typeof(TSource);

    //Create x=>x.PropName
    var propertyInfo = entityType.GetProperty(propertyName);
    ParameterExpression arg = Expression.Parameter(entityType, "x");
    MemberExpression property = Expression.Property(arg, propertyName);
    var selector = Expression.Lambda(property, new ParameterExpression[] { arg });

    //Get System.Linq.Queryable.OrderBy() method.
    var enumarableType = typeof(System.Linq.Queryable);
    var method = enumarableType.GetMethods()
         .Where(m => m.Name == "OrderBy" && m.IsGenericMethodDefinition)
         .Where(m =>
         {
            var parameters = m.GetParameters().ToList();
            //Put more restriction here to ensure selecting the right overload                
            return parameters.Count == 2;//overload that has 2 parameters
         }).Single();
    //The linq's OrderBy<TSource, TKey> has two generic types, which provided here
    MethodInfo genericMethod = method
         .MakeGenericMethod(entityType, propertyInfo.PropertyType);

    /*Call query.OrderBy(selector), with query and selector: x=> x.PropName
      Note that we pass the selector as Expression to the method and we don't compile it.
      By doing so EF can extract "order by" columns and generate SQL for it.*/
    var newQuery = (IOrderedQueryable<TSource>)genericMethod
         .Invoke(genericMethod, new object[] { query, selector });
    return newQuery;
}

```

Now you can call this overload of `OrderBy` like any other overload of it.  

For example:



```
var cheapestItems = _context.Items.OrderBy("Money").Take(10).ToList();

```

Which translates to:



```
SELECT TOP (10)  {coulmn names} FROM  [dbo].[Items] AS [Extent1] 
       ORDER BY [Extent1].[Money] ASC

```

This approach can be used to define all overloads of `OrderBy` and `OrderByDescending` methods to have `string` property selector.